Gonit Vigyan

⚡ Electricity: The Power That Drives the Modern World! ⚡

Electricity is the lifeline of modern society, powering 🔌 homes, 🏫 schools, 🏥 hospitals, and 🏭 industries. It is a controllable and convenient form of energy that makes our daily lives easier. But have you ever wondered what electricity really is? 🤔 How does it flow through an electric circuit? 🔄 What factors regulate the flow of current? ⚙️

In this blog, we will unravel the mysteries of electricity 🔍, understand how it moves through a circuit, and explore the heating effects of electric current 🔥 and its real-world applications. Stay tuned as we illuminate 💡 the fascinating world of electricity! 🚀

📖 Detailed Notes on Electric Current and Circuit

⚡ 12.1 ELECTRIC CURRENT AND CIRCUIT

🔄 Understanding Current Flow

• We are familiar with air currents 🌬️ (moving air) and water currents 💦 (flowing water in rivers).
• Similarly, when electric charge ⚡ flows through a conductor (like a metallic wire 🧵), we say that there is an electric current in the conductor.

🔦 Example: Torch Circuit

• A torch works when cells (battery) are placed correctly 🪫🔋.
• The battery provides electric current which makes the bulb glow 💡.
• The switch acts as a conducting link 🏗️ between the cell and bulb.
• A continuous and closed path 🛤️ through which electric current flows is called an electric circuit.
• If the circuit is broken ❌ (e.g., switch turned off), the current stops flowing, and the bulb does not glow.

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🧲 How Do We Express Electric Current?

• Electric current (I) = The amount of charge (Q) flowing through a particular area in unit time (t).
• Mathematically, $$I = \frac{Q}{t}$$ where,
  • $I$ = Electric Current (in Ampere (A))
  • $Q$ = Charge (in Coulomb (C))
  • $t$ = Time (in Seconds (s))

🔢 SI Unit of Charge and Current

Quantity	SI Unit	Symbol	Value
Electric Charge	Coulomb	$C$	1 electron = $-1.6 \times 10^{-19} C$
Electric Current	Ampere	$A$	$1A = \frac{1C}{1s}$

📏 Small Units of Current

• Milliampere (mA) = $10^{-3} A$
• Microampere (μA) = $10^{-6} A$

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🧭 Conventional Direction of Current

• In metallic wires 🧵, electrons 🚀 move, causing electric current.
• Historically, electric current was assumed as the flow of positive charges ➕.
• Hence, conventionally, the direction of electric current is taken as opposite to the direction of electron flow.

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📏 Measuring Electric Current: The Ammeter

• Ammeter 🧭 is an instrument that measures electric current.
• It is always connected in series 🔗 in a circuit.
• The circuit diagram 📉 consists of:
  • Cell (Battery) 🔋
  • Electric Bulb 💡
  • Ammeter 🧭
  • Plug Key (Switch) 🔑
• Current flows in the circuit from the positive terminal ➕ of the cell to the negative terminal ➖ through the bulb and ammeter.

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📌 Summary

✅ Electric current = Flow of charges in a conductor.
✅ SI Unit of current = Ampere (A) = 1 Coulomb charge passing per second.
✅ Circuit = A closed path where electricity flows.
✅ Ammeter measures current and is connected in series.
✅ Conventional current direction = Opposite to electron flow.

Q U E S T I O N S & A N S W E R S

Example 12.1

Question: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.

Solution:
Given,

• Current (I) = 0.5 A
• Time (t) = 10 min = 600 s

Using the formula:

$$Q = I \times t$$ $$Q = 0.5 A \times 600 s$$ $$Q = 300 C$$

Answer: The total electric charge that flows through the circuit is 300 Coulombs (C). ⚡

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Conceptual Questions

1. What does an electric circuit mean?

Answer:
An electric circuit is a continuous and closed path 🛤️ through which electric current flows. It consists of a power source 🔋 (such as a battery), conducting wires 🧵, and electrical components (like bulbs 💡, resistors, or switches 🔑). If the circuit is broken, the current stops flowing.

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2. Define the unit of current.

Answer:
The SI unit of electric current is the ampere (A), named after Andre-Marie Ampere. It is defined as:

$$1 \text{ Ampere} = \frac{1 \text{ Coulomb}}{1 \text{ Second}}$$

That means 1 ampere is the flow of 1 coulomb of charge per second through a conductor.

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3. Calculate the number of electrons constituting one coulomb of charge.

Answer:
We know that the charge of one electron is:

$$e = 1.6 \times 10^{-19} C$$

The number of electrons (n) in 1 coulomb of charge is:

$$n = \frac{1C}{e}$$ $$n = \frac{1}{1.6 \times 10^{-19}}$$ $$n = 6.25 \times 10^{18} \text{ electrons}$$

So, one coulomb of charge is constituted by approximately $6.25 \times 10^{18}$ electrons. 🔢⚡

📌 Short Notes on ‘Flow’ of Charges Inside a Wire

🔬 How Does a Metal Conduct Electricity?

• Metals conduct electricity due to the presence of free electrons 🔄.
• In a solid metal, atoms are tightly packed, yet electrons can move freely as if in a vacuum.

⚡ Electron Motion in a Conductor

• When an electric current flows, electrons move with an average drift speed 🚶, which is very small (~ 1 mm/s in copper wire).
• The motion of electrons in a wire is not like free-moving charges in space.

💡 Why Does a Bulb Light Up Instantly?

• The bulb glows immediately after switching ON, even though electrons move slowly.
• This happens because electricity travels as an electromagnetic wave ⚡ at a speed close to the speed of light 🌍🚀.
• The actual mechanism of current flow is more complex and requires advanced physics to understand.

📖 Detailed Notes on Electric Potential and Potential Difference

⚡ 12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

🔍 What Makes Electric Charge Flow?

• Charges do not flow in a conductor (e.g., copper wire) on their own, just like water in a horizontal tube does not flow. 💧
• Water flows only when there is a pressure difference between two points (like from a tank at a higher level).
• Similarly, electric charge flows in a conductor only if there is a potential difference (difference in electric pressure).
• Gravity has no role in the movement of electrons in a metallic wire.

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🔋 Role of a Battery in Charge Flow

• A battery 🔋 (or a cell) creates a potential difference in a circuit.
• Chemical action within a cell produces this potential difference, even when no current is drawn.
• When the cell is connected to a circuit, this potential difference moves charges ⚡, producing electric current.
• To maintain the current, the cell uses its stored chemical energy.

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📏 Definition of Electric Potential Difference

• The electric potential difference between two points in a circuit is: $$\text{Potential Difference} (V) = \frac{\text{Work Done} (W)}{\text{Charge} (Q)}$$
• This means potential difference is the work done per unit charge to move a charge between two points.

🔢 SI Unit of Potential Difference

• The SI unit of potential difference is the volt (V), named after Alessandro Volta (1745–1827), an Italian physicist.
• 1 Volt = The potential difference when 1 joule of work is done to move 1 coulomb of charge. $$1 V = \frac{1 J}{1 C}$$
• Therefore, $$1 \text{ Volt} = 1 \text{ Joule per Coulomb} (1V = 1 J C^{-1})$$

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🧭 Measuring Potential Difference: The Voltmeter

• A voltmeter is used to measure the potential difference between two points in a circuit.
• Connection Rule:
  • A voltmeter is always connected in parallel 🔗 across the circuit component whose potential difference is to be measured.

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📌 Summary

✅ Electric charge flows only when there is a potential difference.
✅ A battery (or cell) creates potential difference using chemical energy.
✅ Potential difference is the work done per unit charge to move it between two points.
✅ SI unit of potential difference = Volt (V).
✅ 1V = 1J/C (1 Volt = 1 Joule per Coulomb).
✅ A voltmeter measures potential difference and is connected in parallel.

 

Q U E S T I O N S & A N S W E R S

Example 12.2

Question: How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

Solution:
Given,

• Charge (Q) = 2 C
• Potential Difference (V) = 12 V

Using the formula:

$$W = V \times Q$$

Substituting the values:

$$W = 12 V \times 2 C$$ $$W = 24 J$$

Answer: The work done in moving the charge is 24 Joules (J). ⚡

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Conceptual Questions

1️⃣ Name a device that helps to maintain a potential difference across a conductor.

Answer:
A battery 🔋 (or cell) helps to maintain a potential difference across a conductor by supplying chemical energy to move charges.

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2️⃣ What is meant by saying that the potential difference between two points is 1 V?

Answer:
A potential difference of 1 volt means that 1 joule of work is done to move 1 coulomb of charge between two points in a circuit.
Mathematically,

$$1V = \frac{1J}{1C}$$

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3️⃣ How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:
Using the formula:

$$W = V \times Q$$

Given,

• Potential Difference (V) = 6 V
• Charge (Q) = 1 C (since we need energy per coulomb)

$$W = 6 V \times 1 C$$ $$W = 6 J$$

Answer: Each coulomb of charge gains 6 joules of energy when passing through a 6 V battery. ⚡🔋

📖 12.3 CIRCUIT DIAGRAM

🔍 Understanding Circuit Diagrams

• An electric circuit consists of:
  • Cell/Battery 🔋
  • Plug Key (Switch) 🔑
  • Electrical Components (e.g., bulbs 💡, resistors, meters)
  • Connecting Wires 🧵
• Instead of drawing real components, we use standard symbols to make circuit diagrams simpler and easier to understand.

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📜 Common Circuit Symbols (Table 12.1)

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📌 Summary

✅ A circuit diagram represents an electric circuit using symbols for easy understanding.
✅ Batteries, resistors, bulbs, switches, meters all have standard symbols.
✅ Ammeter is connected in series, while Voltmeter is connected in parallel.

 

📖 Detailed Notes on Ohm’s Law (12.4) 📏

🔍 Understanding the Relationship Between Potential Difference and Current

• Does potential difference (V) affect current (I)? 🤔
• Ohm’s Law helps us understand the relationship between voltage, current, and resistance in a circuit.
• Activity 12.1 helps verify this relationship using a nichrome wire, ammeter, voltmeter, and multiple 1.5V cells.

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⚡ Ohm’s Law: Definition & Formula

🔬 Experiment Findings

• When we increase the number of cells, the readings of voltmeter (V) and ammeter (I) also increase.
• The ratio V/I remains constant ⚖️, which means: $$V \propto I$$
• This means that potential difference (V) is directly proportional to the current (I) in a conductor, provided the temperature remains constant.
• Mathematical Formulation: $$V = IR$$ where:
  • V = Potential Difference (Volts, V) 🔋
  • I = Current (Amperes, A) 🔄
  • R = Resistance (Ohms, Ω) 🌀

📉 V-I Graph for a Nichrome Wire

• The graph shows a straight-line relationship between Voltage (V) and Current (I).
• This confirms Ohm’s Law, where V ∝ I (provided the temperature remains constant).
• Slope of the graph = Resistance (R) of the conductor.

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⚡ Ohm’s Law Graph (V vs. I)

Here is the Voltage (V) vs. Current (I) graph for Ohm’s Law:

• The straight-line relationship confirms that Voltage (V) is directly proportional to Current (I).
• The slope of the line represents resistance (R), which remains constant.

Would you like me to modify it or explain further? 📊⚡

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🛑 Resistance (R) and Its Significance

• Resistance (R) is the property of a conductor to oppose the flow of electric current.
• Formula for Resistance: $$R = \frac{V}{I}$$
• SI Unit of Resistance = Ohm (Ω)
  • 1 Ohm (Ω) = 1 Volt / 1 Ampere
  • If V = 1V and I = 1A, then R = 1Ω.

🌀 Factors Affecting Resistance

• Material of the conductor
• Length and thickness of the wire
• Temperature of the conductor

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⚙️ Effect of Resistance on Current

• From Ohm’s Law: $$I = \frac{V}{R}$$
• If resistance (R) increases, current (I) decreases.
• If resistance (R) decreases, current (I) increases.

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🛠️ Controlling Current with Resistance

• A variable resistor (rheostat) is used to control current without changing the voltage source.
• Used in fans, heaters, dimmers, etc.

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📌 Activity 12.2: Resistance and Different Components

• Setup:
  • Circuit with a nichrome wire, torch bulb, and 10W bulb.
  • An ammeter to measure current flow.
• Observations:
  • Different components show different ammeter readings ⚡.
  • This indicates that different components have different resistance levels.
  • Nichrome wire has moderate resistance 🌀.
  • Torch bulb has higher resistance 💡.
  • 10W bulb has even higher resistance 🔥.

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🎯 Key Takeaways

✅ Ohm’s Law: $V = IR$ 📏
✅ Resistance (R) opposes current flow ⚡.
✅ Higher resistance → Lower current ⬇️🔄.
✅ Variable resistor (Rheostat) controls current 🔄🌀.
✅ Different materials have different resistance levels 🏗️.

📖 Detailed Notes on Factors Affecting Resistance of a Conductor (12.5) 📏

⚡ Activity 12.3: Understanding Resistance in Conducting Wires

🔬 Experimental Observations:

1️⃣ Same material, increased length – Resistance increases when the length of the wire is doubled.
2️⃣ Same material, increased thickness – Resistance decreases when a thicker wire is used.
3️⃣ Different materials, same size – Resistance varies for different materials.

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📊 Factors Affecting Resistance (R)

Resistance ($R$) of a conductor depends on:

1️⃣ Length of the Conductor ($l$)

• Longer wire → More resistance 🔺
• Shorter wire → Less resistance 🔻
• Mathematically, $$R \propto l$$

2️⃣ Cross-sectional Area ($A$)

• Thicker wire → Less resistance 🔻
• Thinner wire → More resistance 🔺
• Mathematically, $$R \propto \frac{1}{A}$$

3️⃣ Nature of Material

• Different materials have different resistivity ($\rho$).
• Metals (Copper, Silver) → Low resistivity → Good conductors ⚡
• Alloys (Nichrome, Constantan) → Higher resistivity → Used in heating devices 🔥
• Insulators (Glass, Rubber) → Very high resistivity → No current flow 🚫

📏 Combined Formula for Resistance

$$R = \rho \frac{l}{A}$$

where:

• $R$ = Resistance (Ω)
• $\rho$ = Resistivity (Ωm) (material property)
• $l$ = Length of conductor (m)
• $A$ = Cross-sectional area (m²)

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🧲 Electrical Resistivity ($\rho$)

• Resistivity is a material-specific constant.
• SI unit: Ohm-meter (Ω·m)
• Lower resistivity → Better conductor ✅
• Higher resistivity → Poor conductor or insulator ❌

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📊 Electrical Resistivity Table at 20°C

🔹 Conductors (Good Electricity Flow) ⚡

Material	Resistivity (Ω·m)
Silver 🥈	$1.60 \times 10^{-8}$
Copper 🟠	$1.62 \times 10^{-8}$
Aluminium ⚪	$2.63 \times 10^{-8}$
Tungsten 💡	$5.20 \times 10^{-8}$
Iron ⚙️	$10.0 \times 10^{-8}$

🔸 Alloys (Higher Resistivity, Used in Heating Devices) 🔥

Alloy	Resistivity (Ω·m)
Constantan (Cu-Ni)	$49 \times 10^{-6}$
Manganin (Cu-Mn-Ni)	$44 \times 10^{-6}$
Nichrome (Ni-Cr-Mn-Fe)	$100 \times 10^{-6}$

🔻 Insulators (Very High Resistivity, No Current Flow) 🚫

Material	Resistivity (Ω·m)
Glass 🪟	$10^{10} - 10^{14}$
Hard Rubber 🔴	$10^{13} - 10^{16}$
Ebonite 🖤	$10^{15} - 10^{17}$
Diamond 💎	$10^{12} - 10^{13}$
Dry Paper 📄	$10^{12}$

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🛠️ Practical Applications of Resistivity

✅ Copper & Aluminium – Used in electrical wiring due to low resistivity.
✅ Nichrome & Manganin – Used in heating devices like toasters, irons.
✅ Tungsten – Used in bulb filaments due to high melting point.
✅ Glass & Rubber – Used as insulators in cables and electrical appliances.

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🎯 Key Takeaways

✅ Resistance increases with length and decreases with thickness.
✅ Material resistivity determines if it’s a conductor, resistor, or insulator.
✅ Formula: $R = \rho \frac{l}{A}$ 📏
✅ Resistivity is a constant for a given material, unlike resistance which depends on size.
✅ Alloys are used in heating devices because they don’t oxidize easily.

📝 Example-Based Questions & Answers

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Example 1: Current Drawn by Electrical Appliances

❓ Question:

(a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?
(b) How much current will an electric heater draw from a 220 V source, if the resistance of the heater coil is 100 Ω?

✅ Solution:

(a) For the Electric Bulb:
Given:

• Voltage (V) = 220 V
• Resistance (R) = 1200 Ω

Using Ohm’s Law:

$$I = \frac{V}{R}$$

Substituting the values:

$$I = \frac{220}{1200} = 0.18 A$$

✅ Current drawn by the bulb = 0.18 A

(b) For the Electric Heater:
Given:

• Voltage (V) = 220 V
• Resistance (R) = 100 Ω

Using Ohm’s Law:

$$I = \frac{V}{R}$$

Substituting the values:

$$I = \frac{220}{100} = 2.2 A$$

✅ Current drawn by the heater = 2.2 A

🔍 Observation:

• The electric heater draws more current than the bulb because it has lower resistance.

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Example 2: Effect of Voltage on Current

❓ Question:

The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source.
What current will the heater draw if the potential difference is increased to 120 V?

✅ Solution:

Given:

• Voltage (V) = 60 V
• Current (I) = 4 A

Using Ohm’s Law to find resistance:

$$R = \frac{V}{I} = \frac{60}{4} = 15 Ω$$

When the potential difference increases to 120 V, the new current is:

$$I = \frac{V}{R} = \frac{120}{15} = 8 A$$

✅ The current through the heater becomes 8 A when the voltage is doubled.

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Example 3: Finding Resistivity of a Metal Wire

❓ Question:

The resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at this temperature?
Using Table 12.2, predict the material of the wire.

✅ Solution:

Given:

• Resistance (R) = 26 Ω
• Diameter (d) = 0.3 mm = $3 \times 10^{-4}$ m
• Length (l) = 1 m

Using the formula for Resistivity ($\rho$):

$$\rho = \frac{R \times A}{l}$$

Since the cross-sectional area of a wire is:

$$A = \frac{\pi d^2}{4}$$

Substituting values:

$$\rho = 1.84 \times 10^{-6} \, \Omega m$$

From Table 12.2, the resistivity matches manganese.
✅ The wire is made of manganese.

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Example 4: Resistance of a New Wire

❓ Question:

A wire of given material having length $l$ and cross-sectional area $A$ has a resistance of 4 Ω.
What would be the resistance of another wire of the same material having length $l/2$ and area of cross-section $2A$?

✅ Solution:

For the first wire:

$$R_1 = \rho \frac{l}{A} = 4 Ω$$

For the second wire:

$$R_2 = \rho \frac{(l/2)}{(2A)}$$ $$R_2 = \frac{1}{4} \times R_1$$ $$R_2 = \frac{1}{4} \times 4Ω = 1Ω$$

✅ The resistance of the new wire is 1 Ω.

📝 Q U E S T I O N S & A N S W E R S

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1️⃣ On what factors does the resistance of a conductor depend?

Answer:
Resistance ($R$) of a conductor depends on:
✅ Length ($l$) – Longer wire → More resistance 📏
✅ Cross-sectional area ($A$) – Thicker wire → Less resistance 🟢
✅ Material of the conductor – Different materials have different resistivity ($\rho$) ⚙️
✅ Temperature – Higher temperature → More resistance 🔥

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2️⃣ Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:
✅ Current flows more easily through a thick wire because resistance is inversely proportional to cross-sectional area ($R \propto 1/A$).
✅ Thicker wire → Less resistance → More current flow.

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3️⃣ Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:
By Ohm’s Law:

$$I = \frac{V}{R}$$

• If V is halved and R remains the same, then:
  ✅ Current will also be halved.

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4️⃣ Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:
✅ Alloys have higher resistivity → More heat generation 🔥
✅ Alloys do not oxidize easily → Longer lifespan ⚙️
✅ Pure metals have lower resistivity → Not ideal for heating devices.

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5️⃣ Use the data in Table 12.2 to answer the following:

(a) Which among iron and mercury is a better conductor?

Answer:
✅ Iron ($10.0 \times 10^{-8} Ωm$) is a better conductor than mercury ($94.0 \times 10^{-8} Ωm$) because it has lower resistivity.

(b) Which material is the best conductor?

Answer:
✅ Silver ($1.60 \times 10^{-8} Ωm$) is the best conductor among all materials.

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📖 12.6 Resistance of a System of Resistors

⚡ Understanding Resistor Combinations

• Resistors can be combined in two ways in an electric circuit:
  1️⃣ Series Combination – Resistors are connected end to end 🔗.
  2️⃣ Parallel Combination – Resistors are connected across common points 🔌.

🔬 Activity 12.4: Resistors in Series

1️⃣ Connect three resistors of different values (e.g., 1Ω, 2Ω, 3Ω) in series with a battery, ammeter, and switch.
2️⃣ Plug in the key and note the ammeter reading.
3️⃣ Move the ammeter to different positions in the circuit and observe the current.

🔍 Observation:
✅ The current remains the same through each resistor.

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12.6.1 Resistors in Series

What happens when resistors are connected in series?
✅ The current remains the same throughout the circuit.
✅ The total potential difference is the sum of individual voltages.
✅ The total resistance is the sum of all resistors.

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🔬 Activity 12.5: Measuring Potential Difference in Series Circuit

1️⃣ Insert a voltmeter across the series combination of resistors.
2️⃣ Measure the potential difference (V) across all resistors.
3️⃣ Measure individual voltage drops across each resistor (V₁, V₂, V₃).
4️⃣ Compare the total voltage (V) with the sum of V₁, V₂, and V₃.

🔍 Observation:
✅ Total Voltage = Sum of Individual Voltages

$$V = V_1 + V_2 + V_3$$

✅ The total resistance of the circuit is:

$$R_s = R_1 + R_2 + R_3$$

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📏 Formula for Resistors in Series

Using Ohm’s Law:

$$V = I R$$

For each resistor:

$$V_1 = I R_1, \quad V_2 = I R_2, \quad V_3 = I R_3$$

Summing these equations:

$$V = V_1 + V_2 + V_3$$

Since current (I) remains the same, we get:

$$I R_s = I (R_1 + R_2 + R_3)$$

Thus,

$$R_s = R_1 + R_2 + R_3$$

🔍 Conclusion:
✅ In a series circuit, the total resistance is the sum of all individual resistances.
✅ The total resistance is always greater than any individual resistance.

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🎯 Key Takeaways

✅ In a series circuit, current remains the same throughout.
✅ Voltage divides across each resistor 🔋.
✅ Total resistance is the sum of all individual resistances.
✅ Series circuits increase overall resistance.

📝 Example 12.7: Resistors in Series

❓ Question:

An electric lamp (resistance 20 Ω) and a conductor (resistance 4 Ω) are connected in series to a 6 V battery. 

Calculate:
(a) The total resistance of the circuit.
(b) The current through the circuit.
(c) The potential difference across the electric lamp and the conductor.

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✅ Solution:

(a) Total Resistance of the Circuit

Since the resistances are connected in series, the total resistance is:

$$R_s = R_1 + R_2$$

Substituting values:

$$R_s = 20Ω + 4Ω = 24Ω$$

✅ Total Resistance = 24 Ω

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(b) Current Through the Circuit

Using Ohm’s Law:

$$I = \frac{V}{R_s}$$

Substituting values:

$$I = \frac{6V}{24Ω}$$ $$I = 0.25A$$

✅ Current Through the Circuit = 0.25 A

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(c) Potential Difference Across the Electric Lamp and Conductor

Using Ohm’s Law ($V = IR$) for each component:

1️⃣ Potential Difference Across the Electric Lamp

$$V_1 = I \times R_1$$ $$V_1 = 0.25A \times 20Ω$$ $$V_1 = 5V$$

✅ Voltage across the lamp = 5 V

2️⃣ Potential Difference Across the Conductor

$$V_2 = I \times R_2$$ $$V_2 = 0.25A \times 4Ω$$ $$V_2 = 1V$$

✅ Voltage across the conductor = 1 V

🔍 Observation:

• Total Voltage (V) = Sum of Individual Voltages: $$V = V_1 + V_2 = 5V + 1V = 6V$$

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🔄 Equivalent Resistance of the Circuit

If we replace the lamp and conductor with an equivalent single resistor, it should have the same total resistance that allows 0.25 A current when connected to 6 V.

Using Ohm’s Law:

$$R = \frac{V}{I}$$

Substituting values:

$$R = \frac{6V}{0.25A}$$ $$R = 24Ω$$

✅ The equivalent resistance is 24 Ω, which matches the total series resistance.

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🎯 Key Takeaways

✅ In a series circuit, total resistance is the sum of individual resistances.
✅ The same current flows through all components.
✅ The total voltage is divided among the resistors according to their resistance.
✅ The equivalent resistance of a series circuit is always greater than any individual resistance.

Q U E S T I O N S & A N S W E R S

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❓ 1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.

📝 Answer:

• The total voltage of the battery = $3 \times 2V = 6V$
• The resistors (5Ω, 8Ω, and 12Ω) are connected in series.
• A plug key (switch) is included in the circuit.

📌 Circuit Diagram:

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❓ 2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

📝 Answer:

• An ammeter is connected in series to measure the current.
• A voltmeter is connected in parallel across the 12Ω resistor to measure the potential difference.

📌 Calculating Current ($I$):
The total resistance in the circuit:

$$R_s = R_1 + R_2 + R_3 = 5Ω + 8Ω + 12Ω = 25Ω$$

Using Ohm’s Law:

$$I = \frac{V}{R_s} = \frac{6V}{25Ω} = 0.24A$$

✅ Ammeter reading = 0.24 A

📌 Calculating Voltage Across the 12Ω Resistor ($V_{12Ω}$):
Using Ohm’s Law:

$$V = I \times R$$ $$V_{12Ω} = 0.24A \times 12Ω = 2.88V$$

✅ Voltmeter reading = 2.88 V

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🎯 Final Answers:

✅ Ammeter Reading = 0.24 A
✅ Voltmeter Reading = 2.88 V

 

📖 12.6.2 Resistors in Parallel

⚡ Understanding Parallel Resistor Combinations

• In a parallel circuit, resistors are connected so that each has the same voltage across it.
• Unlike a series circuit, the total resistance in a parallel circuit is less than any individual resistance.
• Practical Example: Parallel circuits are used in household wiring, where appliances operate independently.

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🔬 Activity 12.6: Studying Resistors in Parallel

1️⃣ Connect three resistors ($R_1, R_2, R_3$) in parallel to a battery, a plug key, and an ammeter as shown in Fig. 12.10.
2️⃣ Connect a voltmeter in parallel with the resistor combination.
3️⃣ Plug in the key and note the ammeter reading ($I$) and the voltmeter reading ($V$).
4️⃣ Measure current ($I_1, I_2, I_3$) through each resistor by placing an ammeter in series with each resistor (see Fig. 12.11).

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📏 Observations & Formula for Parallel Circuits

✅ Total Current is the Sum of Individual Currents

$$I = I_1 + I_2 + I_3$$

✅ Same Voltage Across All Resistors

$$V = V_1 = V_2 = V_3$$

✅ Using Ohm’s Law ($V = IR$) on Each Resistor:

$$I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3}$$

✅ Total Resistance Formula for Parallel Circuits

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

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🎯 Key Takeaways

✅ Voltage remains constant across all parallel resistors.
✅ Total current is the sum of currents through each resistor.
✅ Total resistance in parallel is always less than the smallest resistor.
✅ Parallel circuits allow independent operation of components (e.g., home appliances).

Q U E S T I O N S & A N S W E R S

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❓ Question 1:

In the circuit diagram given in Fig. 12.10, suppose the resistors $R_1 = 5Ω$, $R_2 = 10Ω$, and $R_3 = 30Ω$ are connected in parallel to a 12V battery. Calculate:

(a) The current through each resistor.
(b) The total current in the circuit.
(c) The total resistance in the circuit.

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✅ Answer 1:

(a) Current through each resistor

Using Ohm’s Law:

$$I = \frac{V}{R}$$

For $R_1 = 5Ω$:

$$I_1 = \frac{12V}{5Ω} = 2.4A$$

For $R_2 = 10Ω$:

$$I_2 = \frac{12V}{10Ω} = 1.2A$$

For $R_3 = 30Ω$:

$$I_3 = \frac{12V}{30Ω} = 0.4A$$

✅ Current through $R_1$ = 2.4 A
✅ Current through $R_2$ = 1.2 A
✅ Current through $R_3$ = 0.4 A

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(b) Total Current in the Circuit

$$I = I_1 + I_2 + I_3$$ $$I = (2.4 + 1.2 + 0.4) A = 4 A$$

✅ Total Current = 4 A

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(c) Total Resistance of the Circuit

Using the parallel resistance formula:

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}$$

Converting to a common denominator:

$$\frac{6}{30} + \frac{3}{30} + \frac{1}{30} = \frac{10}{30}$$ $$R_p = \frac{30}{10} = 3Ω$$

✅ Total Resistance = 3 Ω

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❓ Question 2:

In Fig. 12.12, suppose:

• $R_1 = 10Ω$, $R_2 = 40Ω$ (parallel combination)
• $R_3 = 30Ω$, $R_4 = 20Ω$, $R_5 = 60Ω$ (parallel combination)
• A 12V battery is connected to the arrangement.

Find:
(a) The total resistance of the circuit.
(b) The total current flowing in the circuit.

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✅ Answer 2:

(a) Finding Equivalent Resistance

Step 1: Equivalent Resistance of $R_1$ & $R_2$ (Parallel Combination)

$$\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\frac{1}{R'} = \frac{1}{10} + \frac{1}{40} = \frac{4}{40} + \frac{1}{40} = \frac{5}{40}$$ $$R' = \frac{40}{5} = 8Ω$$

✅ Equivalent Resistance of $R_1$ & $R_2$ = 8Ω

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Step 2: Equivalent Resistance of $R_3$, $R_4$, & $R_5$ (Parallel Combination)

$$\frac{1}{R''} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$$ $$\frac{1}{R''} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60}$$

Finding the common denominator (LCM = 60):

$$\frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{6}{60}$$ $$R'' = \frac{60}{6} = 10Ω$$

✅ Equivalent Resistance of $R_3, R_4, R_5$ = 10Ω

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Step 3: Total Resistance in the Circuit

Since $R'$ and $R''$ are in series, total resistance is:

$$R = R' + R'' = 8Ω + 10Ω = 18Ω$$

✅ Total Resistance = 18 Ω

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(b) Total Current in the Circuit

Using Ohm’s Law:

$$I = \frac{V}{R}$$ $$I = \frac{12V}{18Ω} = 0.67 A$$

✅ Total Current = 0.67 A

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🎯 Key Takeaways

✅ In parallel circuits, total resistance is always smaller than the smallest resistor.
✅ Current is divided among parallel branches, but voltage remains the same across each branch.
✅ In a series circuit, total resistance is the sum of individual resistances.
✅ Parallel circuits are more practical for household wiring since each appliance gets the required voltage.

Q U E S T I O N S & A N S W E R S

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❓ 1. Judge the equivalent resistance when the following are connected in parallel:

(a) $1Ω$ and $10^6Ω$
(b) $1Ω$, $10^3Ω$, and $10^6Ω$

✅ Answer:

Using the parallel resistance formula:

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$

(a) For $1Ω$ and $10^6Ω$:

$$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^6}$$ $$\frac{1}{R_p} \approx 1 + 0.000001 = 1.000001$$ $$R_p \approx 1Ω$$

✅ Equivalent resistance = 1Ω

(b) For $1Ω$, $10^3Ω$, and $10^6Ω$:

$$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6}$$ $$\frac{1}{R_p} \approx 1 + 0.001 + 0.000001 = 1.001001$$ $$R_p \approx 0.999Ω \approx 1Ω$$

✅ Equivalent resistance ≈ 1Ω

---

❓ 2. An electric lamp of 100W, a toaster of resistance 50Ω, and a water filter of resistance 500Ω are connected in parallel to a 220V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

✅ Answer:

Step 1: Find total current drawn by the three appliances.

Using Ohm’s Law:

$$I = \frac{V}{R}$$

For Toaster ($R = 50Ω$):

$$I_1 = \frac{220V}{50Ω} = 4.4A$$

For Water Filter ($R = 500Ω$):

$$I_2 = \frac{220V}{500Ω} = 0.44A$$

For Electric Lamp ($P = 100W$):
Using Power Formula:

$$P = VI$$ $$I = \frac{P}{V} = \frac{100W}{220V} = 0.45A$$

Total current:

$$I_{total} = I_1 + I_2 + I_3 = 4.4A + 0.44A + 0.45A = 5.29A$$

Step 2: Find resistance of an electric iron drawing the same current.

Using Ohm’s Law:

$$R = \frac{V}{I}$$ $$R = \frac{220V}{5.29A} = 41.58Ω$$

✅ Resistance of the electric iron = 41.58Ω

✅ Current through it = 5.29A

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❓ 3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

✅ Answer:

✅ Each device gets the same voltage as the power source.
✅ If one device fails, others keep working (unlike series circuits where failure of one breaks the circuit).
✅ Each device operates independently without affecting the others.
✅ More efficient for home appliances, as they can have different power ratings.

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❓ 4. How can three resistors of resistances $2Ω, 3Ω,$ and $6Ω$ be connected to give a total resistance of:

(a) 4Ω
(b) 1Ω

✅ Answer:

(a) To get $R = 4Ω$

• Connect $3Ω$ and $6Ω$ in parallel, then add $2Ω$ in series.
• Parallel resistance formula:

$$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6}$$ $$R_p = \frac{6}{3} = 2Ω$$

• Now, add $2Ω$ in series:

$$R = R_p + R_1 = 2Ω + 2Ω = 4Ω$$

✅ Required connection: $3Ω$ and $6Ω$ in parallel, then in series with $2Ω$.

---

(b) To get $R = 1Ω$

• Connect all three resistors in parallel:

$$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$$ $$\frac{1}{R_p} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$$ $$R_p = 1Ω$$

✅ Required connection: All three resistors in parallel.

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❓ 5. What is:

(a) The highest total resistance
(b) The lowest total resistance
that can be obtained using four resistors: $4Ω, 8Ω, 12Ω,$ and $24Ω$?

✅ Answer:

(a) Highest Resistance

• For maximum resistance, connect all resistors in series.

$$R_s = R_1 + R_2 + R_3 + R_4$$ $$R_s = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω$$

✅ Highest total resistance = 48Ω

---

(b) Lowest Resistance

• For minimum resistance, connect all resistors in parallel.

$$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$$

Finding common denominator (LCM = 24):

$$\frac{6}{24} + \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{12}{24}$$ $$R_p = \frac{24}{12} = 2Ω$$

✅ Lowest total resistance = 2Ω

---

🎯 Summary of Answers:

1️⃣ Equivalent resistance for parallel circuits: $1Ω$ in both cases.
2️⃣ Resistance of iron = $41.58Ω$, Current = $5.29A$.
3️⃣ Parallel circuits provide independent operation & constant voltage.
4️⃣ $4Ω$ by $3Ω$ & $6Ω$ in parallel, then series with $2Ω$; $1Ω$ by all in parallel.
5️⃣ Highest resistance = $48Ω$, Lowest resistance = $2Ω$.

 

🔥 Detailed Notes on the Heating Effect of Electric Current

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📌 12.7 Heating Effect of Electric Current

🔋 Energy Transformation in an Electrical Circuit

• A battery or cell provides electrical energy by maintaining a potential difference across its terminals.
• Electrons flow due to this potential difference, creating an electric current through a resistor or a system of resistors.
• To maintain the current, the battery must continuously expend energy.

⚡ Where Does This Energy Go?

1️⃣ Useful Work: Some energy is used to perform work, such as:

• Rotating a fan blade
• Producing light in a bulb
  2️⃣ Heat Generation: The rest of the energy is converted into heat, which increases the temperature of the device.
• Example: A fan motor heats up after long use
  3️⃣ Purely Resistive Circuits: If a circuit only contains resistors, all the supplied energy is converted into heat.

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🔥 Heating Effect of Electric Current

• The process of converting electrical energy into heat energy in a conductor is called the Heating Effect of Electric Current.
• Devices utilizing this effect:
  • Electric heater
  • Electric iron
  • Electric oven

📜 Joule’s Law of Heating

The heat produced in a resistor is:

$$H = VIt$$

Using Ohm’s Law ($V = IR$):

$$H = I^2 Rt$$

👉 Joule’s Law states that:

• $H$ is directly proportional to $I^2$ (square of the current)
• $H$ is directly proportional to $R$ (resistance of the conductor)
• $H$ is directly proportional to $t$ (time for which current flows)

---

📊 Mathematical Derivation

🔹 Energy Supplied by the Source

• Charge $Q$ flowing in time $t$: $$Q = It$$
• Work done (Energy transferred) to move charge $Q$: $$W = VQ = VIt$$

Thus, the heat energy produced in a resistor is:

$$H = VIt$$

🔹 Using Ohm’s Law ($V = IR$)

Substituting $V = IR$ in $H = VIt$, we get:

$$H = I^2Rt$$

👉 This equation is known as Joule’s Law of Heating.

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💡 Applications of Joule’s Law

1️⃣ Electric Heaters & Geysers

• High resistance coils (nichrome wires) convert electrical energy into heat efficiently.

2️⃣ Electric Bulbs (Filament Lamps)

• Tungsten filament in the bulb gets heated due to the current and glows brightly.

3️⃣ Electric Irons & Water Heaters

• Use high-resistance alloys to generate heat safely.

4️⃣ Electric Fuses

• A thin wire melts when excessive current passes, preventing damage to appliances.

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🎯 Key Takeaways

✅ Heat production is proportional to $I^2$, $R$, and $t$.
✅ Joule’s Law governs the heating effect of current.
✅ Electric appliances like heaters, irons, and bulbs work using this principle.
✅ Ohm’s Law helps calculate heat dissipation.

 

Q U E S T I O N S & A N S W E R S

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❓ 1. An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

✅ Answer:

From Power Formula:

$$P = VI$$

Rearranging for current:

$$I = \frac{P}{V}$$

(a) When heating is at the maximum rate

$$I = \frac{840 W}{220 V} = 3.82 A$$

Now, using Ohm’s Law:

$$R = \frac{V}{I}$$ $$R = \frac{220 V}{3.82 A} = 57.60Ω$$

✅ Current = 3.82 A
✅ Resistance = 57.60 Ω

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(b) When heating is at the minimum rate

$$I = \frac{360 W}{220 V} = 1.64 A$$

Now, using Ohm’s Law:

$$R = \frac{V}{I}$$ $$R = \frac{220 V}{1.64 A} = 134.15Ω$$

✅ Current = 1.64 A
✅ Resistance = 134.15 Ω

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❓ 2. 100 J of heat is produced each second in a 4Ω resistor. Find the potential difference across the resistor.

✅ Answer:

Given Data:

• Heat produced: $H = 100 J$
• Resistance: $R = 4Ω$
• Time: $t = 1s$
• Voltage: $V = ?$

Using Joule’s Law of Heating:

$$H = I^2 R t$$

Rearrange for current $I$:

$$I = \sqrt{\frac{H}{R t}}$$ $$I = \sqrt{\frac{100 J}{4Ω × 1s}}$$ $$I = \sqrt{25} = 5 A$$

Now, using Ohm’s Law:

$$V = IR$$ $$V = 5 A × 4Ω = 20 V$$

✅ Potential Difference = 20 V

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🎯 Key Takeaways:

✔ Power and Resistance Relationship: Higher power means lower resistance.
✔ Joule’s Law of Heating: Helps in determining current when heat is given.
✔ Ohm’s Law Application: Voltage can be found using $V = IR$.

Would you like a graph or circuit diagram for better understanding? ⚡🔥

Q U E S T I O N S & A N S W E R S

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❓ 1. Why does the cord of an electric heater not glow while the heating element does?

✅ Answer:

The cord of an electric heater is usually made of copper or aluminum, which are good conductors with very low resistance.

The heating element, on the other hand, is made of high-resistance materials like nichrome, which resists the flow of current and converts electrical energy into heat energy, making it glow.

✅ Key Reason: The heating element has high resistance, while the cord has low resistance, so most of the electrical energy is converted into heat in the element, causing it to glow.

---

❓ 2. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

✅ Answer:

Given Data:

• Charge, $Q = 96000C$
• Time, $t = 1$ hour = $3600s$
• Voltage, $V = 50V$

Using the formula for electrical energy (heat generated):

$$H = VQ$$

Substituting the values:

$$H = 50V \times 96000C$$ $$H = 4.8 \times 10^6 J = 4.8 MJ$$

✅ Heat Generated = 4.8 MJ (Mega Joules)

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❓ 3. An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.

✅ Answer:

Given Data:

• Resistance, $R = 20Ω$
• Current, $I = 5A$
• Time, $t = 30s$

Using Joule’s Law of Heating:

$$H = I^2 R t$$

Substituting values:

$$H = (5A)^2 \times 20Ω \times 30s$$ $$H = 25 \times 20 \times 30$$ $$H = 15000 J = 15 kJ$$

✅ Heat Developed = 15 kJ (Kilojoules)

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🎯 Summary of Answers:

1️⃣ The heater's cord does not glow because it has low resistance, while the heating element has high resistance.
2️⃣ Heat generated when transferring 96000C charge = 4.8 MJ.
3️⃣ Heat developed in the electric iron in 30s = 15 kJ.

📖 Detailed Notes on Practical Applications of Heating Effect of Electric Current

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🔥 12.7.1 Practical Applications of Heating Effect of Electric Current

⚡ Unavoidable Heating in Electric Circuits

• Heat generation in conductors is a natural consequence of electric current.
• In some cases, it is undesirable as it can:
  • Waste electrical energy.
  • Overheat components, altering their properties.

However, the heating effect is used in many useful applications.

---

🔌 Devices Using Joule’s Heating Effect

✅ Household Electrical Appliances

The heating effect of current is used in:
1️⃣ Electric iron – For pressing clothes.
2️⃣ Electric toaster – For heating and toasting bread.
3️⃣ Electric oven – For baking and cooking food.
4️⃣ Electric kettle – For boiling water and making beverages.
5️⃣ Electric heater – For warming rooms in cold weather.

💡 These appliances contain a high-resistance wire (nichrome wire) that heats up when current passes through it.

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💡 Electric Bulb (Incandescent Lamp)

How Does It Work?

• Filament Material: Tungsten (high melting point 3380°C).
• Why Tungsten? It can withstand high temperatures without melting.
• Gases Inside Bulb: Nitrogen & Argon
  • Prevents tungsten from oxidizing.
  • Increases filament lifespan.
• Energy Conversion:
  • Most energy → Heat
  • Small part → Light (radiation)
• Working Principle: The filament heats up due to high resistance and glows to produce light.

---

🛑 Electric Fuse – Safety Device

📌 Purpose of a Fuse

• Prevents excess current flow that may damage appliances.
• Placed in series with the device to stop overloading.
• Made of materials with low melting points like:
  • Aluminum
  • Copper
  • Iron
  • Lead

📜 How Does It Work?

• If current exceeds the rated value, the fuse wire melts.
• This breaks the circuit and prevents damage.
• The fuse wire is enclosed in a cartridge of porcelain or insulating material.

---

🧮 Fuse Ratings & Example Calculation

Common Fuse Ratings: 1A, 2A, 3A, 5A, 10A, etc.

Example: Fuse Selection for an Electric Iron

• Power of iron: 1 kW = 1000 W
• Voltage supply: 220 V
• Current Calculation: $$I = \frac{P}{V} = \frac{1000}{220} = 4.54A$$
• Appropriate fuse rating: 5A (next available rating).

---

🎯 Key Takeaways

✅ Heating effect of current is used in household appliances like irons, toasters, ovens, kettles, and heaters.
✅ Tungsten is used in bulbs due to its high melting point, and inert gases prolong filament life.
✅ A fuse prevents excessive current from damaging appliances by melting and breaking the circuit.
✅ Fuse ratings should be chosen based on appliance power and voltage supply.

⚡ Detailed Notes on Electric Power & Energy

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📌 12.8 Electric Power

🔋 Definition of Electric Power

• Power is the rate of doing work or rate of energy consumption.
• In electric circuits, power is the rate at which electrical energy is dissipated or used.
• Mathematically, electric power is given by:

$$P = VI$$

Using Ohm’s Law ($V = IR$), alternative formulas are:

$$P = I^2 R$$ $$P = \frac{V^2}{R}$$

⚡ SI Unit of Electric Power

• The SI unit of power is the watt (W).
• 1 watt is the power consumed when 1 ampere of current flows at 1 volt potential difference.

$$1 W = 1 V \times 1 A = 1 VA$$

📏 Larger Unit: Kilowatt (kW)

Since 1 watt is a small unit, we use:

$$1 \text{ kW} = 1000 W$$

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🔌 Electric Energy & Its Units

📜 Definition of Electric Energy

• Electric energy is the total work done by an electric current in a given time.
• Mathematically,

$$\text{Electric Energy} = \text{Power} \times \text{Time}$$

📏 Unit: Watt-hour (Wh) & Kilowatt-hour (kWh)

1. 1 Watt-hour (Wh) = Energy used when 1 W power is used for 1 hour.
2. 1 Kilowatt-hour (kWh) = 1000 W used for 1 hour.
3. 1 kWh = 3.6 × 10⁶ Joules (J)

👉 Commercial electricity is measured in "units". 1 Unit = 1 kWh

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💡 Important Concept: Electrons Are Not Consumed

• Common Misconception: People think that electrons are used up in an electric circuit.
• Reality: Electrons are not consumed; they simply carry energy from the power source to the device.
• What do we pay for? We pay for the electrical energy used, not the electrons themselves.

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🎯 Key Takeaways

✅ Electric power is the rate of energy consumption and is given by $P = VI$, $P = I^2 R$, or $P = V^2 / R$.
✅ SI unit of power is watt (W), and 1 kW = 1000 W.
✅ Electric energy is measured in watt-hour (Wh) and kilowatt-hour (kWh), where 1 kWh = 3.6 × 10⁶ J.
✅ Electricity bills are based on energy consumption in kWh (units), not electrons used.

 

Q U E S T I O N S & A N S W E R S

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❓ 1. An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?

✅ Answer:

Using the Power Formula:

$$P = VI$$

Substituting the given values:

$$P = 220V \times 0.50A$$ $$P = 110 W$$

✅ Power of the bulb = 110 W

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❓ 2. An electric refrigerator rated 400 W operates for 8 hours per day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kWh?

✅ Answer:

Step 1: Calculate Total Energy Consumption

Energy consumed per day:

$$400 W \times 8 \text{ hours} = 3200 Wh = 3.2 kWh$$

Energy consumed in 30 days:

$$3.2 kWh \times 30 = 96 kWh$$

Step 2: Calculate Total Cost

$$\text{Total cost} = 96 kWh \times Rs 3.00 \text{ per kWh}$$ $$= Rs 288.00$$

✅ Cost to operate the refrigerator for 30 days = Rs 288.00

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🎯 Key Takeaways:

✔ Power of a device is calculated using $P = VI$.
✔ Energy consumption is found using $\text{Power} \times \text{Time}$.
✔ Electricity cost depends on energy consumed in kWh and the rate per unit.

Q U E S T I O N S & A N S W E R S

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❓ 1. What determines the rate at which energy is delivered by a current?

✅ Answer:

• The rate at which energy is delivered by a current is known as electric power.
• Electric power (P) depends on:
  • Voltage (V) applied across the circuit.
  • Current (I) flowing through the circuit.
  • Resistance (R) of the circuit.
• It is given by the formula: $$P = VI$$ or using Ohm’s Law: $$P = I^2 R$$ $$P = \frac{V^2}{R}$$

✅ Conclusion: The rate of energy delivery is determined by the voltage and current in the circuit.

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❓ 2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hours.

✅ Answer:

Step 1: Calculate Power

Using Power Formula:

$$P = VI$$ $$P = 220V \times 5A$$ $$P = 1100 W = 1.1 kW$$

Step 2: Calculate Energy Consumed in 2 Hours

$$\text{Energy} = \text{Power} \times \text{Time}$$ $$= 1.1 \text{ kW} \times 2 \text{ h}$$ $$= 2.2 \text{ kWh}$$

✅ Power of the motor = 1100 W (1.1 kW)
✅ Energy consumed in 2 hours = 2.2 kWh (units)

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🎯 Key Takeaways:

✔ Power depends on voltage and current ($P = VI$).
✔ Energy consumed is calculated using $\text{Energy} = P \times t$.
✔ Electricity is billed based on energy consumption in kilowatt-hours (kWh).

E X E R C I S E S

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❓ 1. A piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R'$, then the ratio $\frac{R}{R'}$ is –

(a) $\frac{1}{25}$
(b) $\frac{1}{5}$
(c) 5
(d) 25

✅ Answer:

Each part of the wire has resistance $R' = \frac{R}{5}$.
Since they are connected in parallel, the total resistance is:

$$\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$$ $$\frac{1}{R'} = 5 \times \frac{1}{\left( \frac{R}{5} \right)} = \frac{5}{R} \times 5 = \frac{25}{R}$$ $$R' = \frac{R}{25}$$

Thus,

$$\frac{R}{R'} = 25$$

✅ Correct answer: (d) 25

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❓ 2. Which of the following terms does not represent electrical power in a circuit?

(a) $I^2 R$
(b) $IR^2$
(c) $VI$
(d) $V^2 / R$

✅ Answer:

The correct formulas for electric power are:

$$P = VI$$ $$P = I^2 R$$ $$P = \frac{V^2}{R}$$

The term $IR^2$ is incorrect.

✅ Correct answer: (b) $IR^2$

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❓ 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

✅ Answer:

Given,

$$P_1 = 100 W, V_1 = 220 V, V_2 = 110 V$$

Using the power formula:

$$P_2 = P_1 \times \left( \frac{V_2}{V_1} \right)^2$$ $$P_2 = 100 \times \left( \frac{110}{220} \right)^2$$ $$P_2 = 100 \times \left( \frac{1}{4} \right)$$ $$P_2 = 25 W$$

✅ Correct answer: (d) 25 W

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❓ 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

✅ Answer:

Using Joule’s Law of Heating:

$$H = I^2 R t$$

For series connection, the total resistance:

$$R_s = R + R = 2R$$

For parallel connection, the total resistance:

$$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$$ $$R_p = \frac{R}{2}$$

Since voltage is constant,

$$H_s = \frac{V^2}{2R} t$$ $$H_p = \frac{V^2}{(R/2)} t = \frac{4V^2}{4R} t$$ $$H_s : H_p = 1 : 4$$

✅ Correct answer: (c) 1:4

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❓ 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

✅ Answer:

A voltmeter is always connected in parallel with the component across which the potential difference is to be measured. This ensures that it measures the voltage drop without drawing significant current.

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❓ 6. A copper wire has diameter 0.5 mm and resistivity of $1.6 \times 10^{-8} \Omega m$. What will be the length of this wire to make its resistance 10 $\Omega$? How much does the resistance change if the diameter is doubled?

✅ Answer:

Given Data:

• Diameter, $d = 0.5 mm = 0.0005 m$
• Radius, $r = \frac{d}{2} = 0.00025 m$
• Resistivity, $\rho = 1.6 \times 10^{-8} \Omega m$
• Resistance, $R = 10 \Omega$

Using Resistance Formula:

$$R = \rho \frac{l}{A}$$

Area of cross-section:

$$A = \pi r^2 = \pi (0.00025)^2$$ $$A = 1.96 \times 10^{-7} m^2$$

Solving for $l$:

$$10 = \frac{(1.6 \times 10^{-8}) \times l}{1.96 \times 10^{-7}}$$ $$l = \frac{10 \times 1.96 \times 10^{-7}}{1.6 \times 10^{-8}}$$ $$l = 122.5 m$$

✅ Length of wire required = 122.5 m

If the diameter is doubled:

• New radius, $r' = 2r = 0.0005 m$
• New area, $A' = \pi (0.0005)^2 = 7.85 \times 10^{-7} m^2$
• New resistance,

$$R' = \rho \frac{l}{A'}$$ $$R' = \frac{(1.6 \times 10^{-8}) \times 122.5}{7.85 \times 10^{-7}}$$ $$R' = 2.5 \Omega$$

✅ New resistance when diameter is doubled = 2.5 $\Omega$
✅ Resistance is reduced to one-fourth of the original value.

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❓ 7. The values of current $I$ flowing in a given resistor for the corresponding values of potential difference $V$ across the resistor are given below –

I (A)	0.5	1.0	2.0	3.0	4.0
V (V)	1.6	3.4	6.7	10.2	13.2

Plot a graph between $V$ and $I$ and calculate the resistance of that resistor.

✅ Answer:

• Plot V on Y-axis and I on X-axis.
• The graph will be a straight line, showing Ohm’s Law.
• Slope of the line = Resistance (R)

$$R = \frac{V}{I}$$

Using any data point,

$$R = \frac{3.4V}{1.0A} = 3.4 \Omega$$

✅ Resistance of the resistor = 3.4 $\mathrm{\Omega }$

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❓ 8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

✅ Answer:

Using Ohm’s Law:

$$R = \frac{V}{I}$$

Given:

• Voltage, $V = 12V$
• Current, $I = 2.5 mA = 2.5 \times 10^{-3} A$

Substituting values:

$$R = \frac{12V}{2.5 \times 10^{-3} A}$$ $$R = 4800 \Omega = 4.8 k\Omega$$

✅ Resistance = 4.8 kΩ

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❓ 9. A battery of 9 V is connected in series with resistors of $0.2 \Omega$, $0.3 \Omega$, $0.4 \Omega$, $0.5 \Omega$, and $12 \Omega$, respectively. How much current would flow through the $12 \Omega$ resistor?

✅ Answer:

Since all resistors are in series, the total resistance is:

$$R_{\text{total}} = 0.2 + 0.3 + 0.4 + 0.5 + 12$$ $$R_{\text{total}} = 13.4 \Omega$$

Using Ohm’s Law:

$$I = \frac{V}{R_{\text{total}}}$$ $$I = \frac{9V}{13.4 \Omega}$$ $$I = 0.67 A$$

✅ Current through the $12 \Omega$ resistor = 0.67 A

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❓ 10. How many $176 \Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

✅ Answer:

Using Ohm’s Law to find total resistance:

$$R_{\text{total}} = \frac{V}{I} = \frac{220V}{5A} = 44 \Omega$$

For parallel resistors:

$$\frac{1}{R_{\text{total}}} = \frac{n}{R}$$ $$\frac{1}{44} = \frac{n}{176}$$ $$n = \frac{176}{44} = 4$$

✅ Number of resistors required = 4

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❓ 11. Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of (i) $9 \Omega$, (ii) $4 \Omega$.

✅ Answer:

(i) To get $9 \Omega$

• Connect two resistors in parallel:

$$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$ $$R_p = 3 \Omega$$

• Connect the third resistor in series:

$$R_{\text{total}} = R_p + R_3 = 3 + 6 = 9 \Omega$$

✅ Two in parallel, one in series → $9 \Omega$

(ii) To get $4 \Omega$

• Connect all three resistors in parallel:

$$\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$ $$R_{\text{total}} = 2 \Omega$$

• Connect this parallel combination in series with another 6 Ω resistor:

$$R_{\text{total}} = 2 + 2 = 4 \Omega$$

✅ Three in parallel → $2 \Omega$, then one in series → $4 \Omega$

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❓ 12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of a 220 V line if the maximum allowable current is 5 A?

✅ Answer:

Using Power Formula:

$$P = VI$$

Total allowable power:

$$P_{\text{total}} = 220V \times 5A = 1100 W$$

For one bulb:

$$P = 10W$$

Number of bulbs:

$$n = \frac{P_{\text{total}}}{P_{\text{one bulb}}} = \frac{1100}{10} = 110$$

✅ Maximum bulbs = 110

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❓ 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils $A$ and $B$, each of $24 \Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

✅ Answer:

Using Ohm’s Law ($I = \frac{V}{R}$):

(i) When used separately

$$I_A = \frac{220V}{24 \Omega} = 9.17A$$

(ii) When used in series

$$R_s = 24 + 24 = 48 \Omega$$ $$I_s = \frac{220V}{48 \Omega} = 4.58A$$

(iii) When used in parallel

$$\frac{1}{R_p} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12}$$ $$R_p = 12 \Omega$$ $$I_p = \frac{220V}{12 \Omega} = 18.33A$$

✅ Current when used separately = 9.17A
✅ Current when used in series = 4.58A
✅ Current when used in parallel = 18.33A

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❓ 14. Compare the power used in the $2 \Omega$ resistor in each of the following circuits:

(i) A 6 V battery in series with $1 \Omega$ and $2 \Omega$ resistors.
(ii) A 4 V battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.

✅ Answer:

(i) In Series

$$R_{\text{total}} = 1 + 2 = 3 \Omega$$ $$I = \frac{V}{R_{\text{total}}} = \frac{6V}{3 \Omega} = 2A$$

Power in $2 \Omega$:

$$P = I^2 R = (2A)^2 \times 2 \Omega = 8W$$

(ii) In Parallel

Voltage across $2 \Omega$ is 4V (same in parallel).

$$P = \frac{V^2}{R} = \frac{4^2}{2} = 8W$$

✅ Power in both cases = 8W

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🎯 Key Takeaways:

✔ Electric power and resistance calculations use Ohm’s Law & Power formulas.
✔ Series and parallel circuits affect total resistance differently.
✔ Fuse selection and current distribution depend on power ratings.

 

E X E R C I S E S (Continued)

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❓ 15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

✅ Answer:

Given:

• Power of Lamp 1 = 100 W
• Power of Lamp 2 = 60 W
• Voltage = 220 V

Using Power Formula:

$$P = VI \Rightarrow I = \frac{P}{V}$$

For Lamp 1:

$$I_1 = \frac{100 W}{220 V} = 0.4545 A$$

For Lamp 2:

$$I_2 = \frac{60 W}{220 V} = 0.2727 A$$

Total current drawn:

$$I_{\text{total}} = I_1 + I_2 = 0.4545 + 0.2727 = 0.727 A$$

✅ Total current drawn = 0.727 A

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❓ 16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?

✅ Answer:

Using Energy Formula:

$$E = P \times t$$

For TV set:

$$E_{\text{TV}} = 250 W \times 1 h = 250 Wh = 0.25 kWh$$

For Toaster:

$$E_{\text{Toaster}} = 1200 W \times \frac{10}{60} h$$ $$E_{\text{Toaster}} = 1200 \times \frac{1}{6} = 200 Wh = 0.2 kWh$$

✅ TV consumes more energy (0.25 kWh) than the toaster (0.2 kWh).

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❓ 17. An electric heater of resistance $8 \Omega$ draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

✅ Answer:

Using Joule’s Law of Heating:

$$H = I^2 R t$$

Given:

• $R = 8 \Omega$, $I = 15 A$, $t = 2 h$

Rate of heat development (Power):

$$P = I^2 R$$ $$P = (15)^2 \times 8$$ $$P = 225 \times 8 = 1800 W = 1.8 kW$$

✅ Rate of heat development = 1.8 kW

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❓ 18. Explain the following:

(a) Why is tungsten used almost exclusively for the filament of electric lamps?

✅ Answer:

• Tungsten has a very high melting point (3380°C), which prevents it from melting under extreme heat.
• It has high resistivity, which allows it to become hot and emit light efficiently.
• It does not oxidize easily when enclosed in an inert gas like argon or nitrogen.

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(b) Why are the conductors of electric heating devices, such as bread toasters and electric irons, made of an alloy rather than a pure metal?

✅ Answer:

• Alloys have higher resistivity than pure metals, allowing them to heat up more efficiently.
• They do not oxidize or burn at high temperatures.
• They offer better durability and last longer.

Examples: Nichrome (Nickel + Chromium + Iron + Manganese) is commonly used.

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(c) Why is the series arrangement not used for domestic circuits?

✅ Answer:

• In a series circuit, the same current flows through all appliances, which is not suitable for devices with different power requirements.
• If one appliance stops working (e.g., a bulb fuses), the entire circuit is broken.
• In a parallel circuit, each device receives the same voltage, ensuring independent operation.

✅ That’s why domestic circuits use parallel arrangements.

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(d) How does the resistance of a wire vary with its area of cross-section?

✅ Answer:

Resistance is inversely proportional to the cross-sectional area ($A$) of a wire:

$$R \propto \frac{1}{A}$$

• If the area increases, resistance decreases.
• If the area decreases (thin wire), resistance increases.

This is why thicker wires are used for high-power appliances to reduce resistance.

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(e) Why are copper and aluminum wires usually employed for electricity transmission?

✅ Answer:

• Copper and aluminum have very low resistivity, making them excellent conductors.
• They have high tensile strength, which helps in overhead wiring.
• Aluminum is lightweight and cheaper than copper, making it ideal for long-distance transmission.

✅ Copper is used in household wiring, while aluminum is used for power transmission lines.

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🎯 Summary of Key Concepts:

✔ Electric power and energy calculations depend on power rating, voltage, and current.
✔ Parallel circuits are preferred for household wiring.
✔ Tungsten is used in bulbs due to its high melting point.
✔ Alloys are used in heating devices for durability and higher resistivity.
✔ Thicker wires have lower resistance, making them ideal for power transmission.