Q1. What is electric current?

Answer: Electric current is the rate of flow of electric charge through a conductor.

I = Q / t

Q2. What is the SI unit of electric current?

Answer: The SI unit of electric current is ampere, denoted by A.

Q3. Define one ampere.

Answer: If one coulomb of charge flows through a conductor in one second, the current is called one ampere.

1 A = 1 C / 1 s

Q4. What is potential difference?

Answer: Potential difference between two points is the work done in moving a unit charge from one point to another.

V = W / Q

Q5. Define one volt.

Answer: Potential difference is one volt if one joule of work is done to move one coulomb of charge.

1 V = 1 J / 1 C

Q6. State Ohm’s law.

Answer: At constant temperature, the current through a conductor is directly proportional to the potential difference across its ends.

V = IR

Q7. What is resistance?

Answer: Resistance is the property of a conductor by which it opposes the flow of electric current.

Q8. What is the SI unit of resistance?

Answer: The SI unit of resistance is ohm, denoted by Ω.

Q9. What is resistivity?

Answer: Resistivity is the resistance of a conductor of unit length and unit area of cross-section. It is a material property.

Q10. What is the SI unit of resistivity?

Answer: The SI unit of resistivity is ohm metre, written as Ω m.

Q11. What is Joule’s heating law?

Answer: The heat produced in a resistor is directly proportional to the square of current, resistance and time.

H = I²Rt

Q12. What is electric power?

Answer: Electric power is the rate at which electrical energy is consumed or work is done.

P = VI

Q13. What is the SI unit of electric power?

Answer: The SI unit of electric power is watt, denoted by W.

Q14. What is the commercial unit of electrical energy?

Answer: The commercial unit of electrical energy is kilowatt hour, written as kWh.

Q15. Write the relation between kWh and joule.

Answer:

1 kWh = 3.6 × 10⁶ J

Q1. Why is an ammeter connected in series?

Answer: An ammeter measures the current flowing through a circuit. Since current is the same in series, an ammeter is connected in series. It has very low resistance so that it does not change the circuit current significantly.

Q2. Why is a voltmeter connected in parallel?

Answer: A voltmeter measures potential difference between two points. Therefore, it is connected in parallel across the component. It has very high resistance so that very little current flows through it.

Q3. What are the conditions for Ohm’s law to be valid?

Answer:

• The temperature of the conductor should remain constant.
• The physical condition of the conductor should not change.
• The conductor should be ohmic, meaning V-I graph should be a straight line.

Q4. What does the slope of a V-I graph represent?

Answer: In a graph of potential difference V on the y-axis and current I on the x-axis, the slope represents resistance.

Slope = V / I = R

Q5. Write the factors on which resistance of a conductor depends.

Answer: Resistance of a conductor depends on:

• Length of the conductor: R ∝ l
• Area of cross-section: R ∝ 1/A
• Nature of material
• Temperature of the conductor

Q6. Why are copper and aluminium used for electrical transmission wires?

Answer: Copper and aluminium have low resistivity, so they offer less resistance to current. This reduces energy loss as heat and makes them suitable for electrical transmission wires.

Q7. Why is tungsten used as filament in electric bulbs?

Answer: Tungsten is used as filament because it has high melting point and high resistance. It can become white hot and emit light without melting easily.

Q8. Why is nichrome used in electric heaters?

Answer: Nichrome is used in electric heaters because it has high resistivity, high melting point and does not oxidise easily at high temperature.

Q9. Why is parallel combination preferred in household wiring?

Answer:

• Each appliance gets the same voltage.
• Each appliance can be switched on or off independently.
• If one appliance fails, others continue to work.
• Equivalent resistance decreases, allowing proper current distribution.

Q10. What is the heating effect of electric current?

Answer: When electric current flows through a resistor, electrical energy is converted into heat energy. This is called heating effect of electric current.

Q11. Write two applications of heating effect of current.

Answer:

• Electric heater, electric iron, geyser and toaster use heating effect of current.
• Electric fuse works on the heating effect of current.

Q12. Why does an electric fuse melt during overloading?

Answer: During overloading, large current flows through the fuse wire. Due to Joule heating, the fuse wire becomes hot and melts, breaking the circuit and protecting appliances.

Q1. Explain Ohm’s law with graph and formula.

Answer: Ohm’s law states that at constant temperature, current flowing through a conductor is directly proportional to the potential difference across its ends.

V ∝ I

V = IR

Here, R is resistance of the conductor.

• If potential difference increases, current also increases proportionally.
• The V-I graph for an ohmic conductor is a straight line passing through the origin.
• The slope of the V-I graph gives resistance.

Q2. Explain resistance and resistivity. How are they different?

Answer: Resistance is the opposition offered by a conductor to the flow of current. Resistivity is a property of the material of the conductor.

R = ρl / A

Q3. Derive equivalent resistance for resistors connected in series.

Answer: In series combination, resistors are connected end to end and the same current flows through each resistor.

Let three resistors R₁, R₂ and R₃ be connected in series.

V = V₁ + V₂ + V₃

Using Ohm’s law:

IR = IR₁ + IR₂ + IR₃

R = R₁ + R₂ + R₃

Conclusion: Equivalent resistance in series is greater than each individual resistance.

Q4. Derive equivalent resistance for resistors connected in parallel.

Answer: In parallel combination, the potential difference across each resistor is the same, but current divides among the branches.

Let three resistors R₁, R₂ and R₃ be connected in parallel.

I = I₁ + I₂ + I₃

Using Ohm’s law:

V/R = V/R₁ + V/R₂ + V/R₃

1/R = 1/R₁ + 1/R₂ + 1/R₃

Conclusion: Equivalent resistance in parallel is less than the smallest individual resistance.

Q5. Compare series and parallel combination of resistors.

Q6. Explain Joule’s heating law and its applications.

Answer: Joule’s heating law states that heat produced in a resistor is:

• Directly proportional to the square of current.
• Directly proportional to resistance.
• Directly proportional to time for which current flows.

H = I²Rt

Applications:

• Electric iron converts electrical energy into heat energy.
• Electric heater, toaster and geyser use heating effect of current.
• Electric bulb filament becomes white hot and produces light.
• Electric fuse melts when excessive current flows and protects the circuit.

Q7. Explain electric power and commercial unit of electrical energy.

Answer: Electric power is the rate at which electrical energy is consumed or work is done.

P = VI

Using Ohm’s law, power can also be written as:

P = I²R

P = V² / R

Electrical energy consumed is:

E = Pt

The commercial unit of electrical energy is kilowatt hour.

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

Q1. A current of 0.5 A flows through a resistor when a potential difference of 10 V is applied. Find resistance.

Solution:

Given: V = 10 V, I = 0.5 A

R = V/I = 10/0.5 = 20 Ω

Answer: Resistance is 20 Ω.

Q2. A wire has resistance 12 Ω. What potential difference is required to pass 2.5 A current through it?

Solution:

Given: R = 12 Ω, I = 2.5 A

V = IR = 2.5 × 12 = 30 V

Answer: Required potential difference is 30 V.

Q3. A charge of 900 C flows through a wire in 5 minutes. Find current.

Solution:

Given: Q = 900 C, t = 5 min = 300 s

I = Q/t = 900/300 = 3 A

Answer: Current is 3 A.

Q4. A conductor has resistance 5 Ω. How much charge flows through it in 2 minutes when connected to a 10 V battery?

Solution:

Given: R = 5 Ω, V = 10 V, t = 2 min = 120 s

I = V/R = 10/5 = 2 A

Q = It = 2 × 120 = 240 C

Answer: Charge flowing is 240 C.

Q1. A wire of length 2 m and area of cross-section 1 × 10⁻⁶ m² has resistance 4 Ω. Find its resistivity.

Solution:

Given: R = 4 Ω, l = 2 m, A = 1 × 10⁻⁶ m²

ρ = RA/l

ρ = 4 × 1 × 10⁻⁶ / 2

ρ = 2 × 10⁻⁶ Ω m

Answer: Resistivity is 2 × 10⁻⁶ Ω m.

Q2. A wire has resistivity 1.6 × 10⁻⁸ Ω m, length 5 m and area 2 × 10⁻⁶ m². Find its resistance.

Solution:

R = ρl/A

R = (1.6 × 10⁻⁸ × 5) / (2 × 10⁻⁶)

R = 4 × 10⁻² Ω = 0.04 Ω

Answer: Resistance is 0.04 Ω.

Q3. A wire is stretched to double its original length. If its original resistance was 6 Ω, find the new resistance assuming volume remains constant.

Solution:

If length is doubled, area becomes half because volume remains constant.

R = ρl/A

New length = 2l and new area = A/2

R’ = ρ(2l)/(A/2) = 4ρl/A = 4R

R’ = 4 × 6 = 24 Ω

Answer: New resistance is 24 Ω.

Q1. Three resistors 2 Ω, 3 Ω and 5 Ω are connected in series with a 20 V battery. Find equivalent resistance and current.

Solution:

R = 2 + 3 + 5 = 10 Ω

I = V/R = 20/10 = 2 A

Answer: Equivalent resistance is 10 Ω and current is 2 A.

Q2. Resistors 6 Ω and 3 Ω are connected in parallel. Find equivalent resistance.

Solution:

1/R = 1/6 + 1/3

1/R = 1/6 + 2/6 = 3/6 = 1/2

R = 2 Ω

Answer: Equivalent resistance is 2 Ω.

Q3. Two resistors 4 Ω and 12 Ω are connected in parallel across a 24 V battery. Find current through each resistor and total current.

Solution:

In parallel, voltage across each resistor is 24 V.

I₁ = 24/4 = 6 A

I₂ = 24/12 = 2 A

I = I₁ + I₂ = 6 + 2 = 8 A

Answer: Current through 4 Ω is 6 A, through 12 Ω is 2 A and total current is 8 A.

Q4. A 3 Ω resistor is connected in series with a parallel combination of 6 Ω and 12 Ω. Find total resistance.

Solution:

First find parallel resistance:

1/R_p = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4

R_p = 4 Ω

Total resistance:

R = 3 + 4 = 7 Ω

Answer: Total resistance is 7 Ω.

Q5. A circuit contains 2 Ω and 4 Ω in series. This combination is connected in parallel with 3 Ω. Find equivalent resistance.

Solution:

Series combination:

R_s = 2 + 4 = 6 Ω

Now 6 Ω is in parallel with 3 Ω:

1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

R = 2 Ω

Answer: Equivalent resistance is 2 Ω.

Q1. A 5 Ω resistor is connected in series with a parallel combination of 10 Ω and 15 Ω. The circuit is connected to a 22 V battery. Find total resistance, total current and current through each parallel branch.

Solution:

Parallel part:

1/R_p = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6

R_p = 6 Ω

Total resistance:

R = 5 + 6 = 11 Ω

Total current:

I = V/R = 22/11 = 2 A

Voltage across 5 Ω resistor:

V₅ = IR = 2 × 5 = 10 V

Voltage across parallel combination:

V_p = 22 − 10 = 12 V

Current through 10 Ω:

I₁₀ = 12/10 = 1.2 A

Current through 15 Ω:

I₁₅ = 12/15 = 0.8 A

Answer: Total resistance = 11 Ω, total current = 2 A, currents are 1.2 A and 0.8 A.

Q2. A 100 W, 220 V bulb and a 60 W, 220 V bulb are connected in series across 220 V. Which bulb glows brighter?

Solution:

Resistance of 100 W bulb:

R₁₀₀ = V²/P = 220²/100 = 484 Ω

Resistance of 60 W bulb:

R₆₀ = 220²/60 = 806.7 Ω

In series, same current flows through both bulbs. Heat/power in each bulb is:

P = I²R

The 60 W bulb has higher resistance, so it consumes more power in series and glows brighter.

Answer: The 60 W bulb glows brighter when connected in series.

Q3. A 100 W, 220 V bulb and a 60 W, 220 V bulb are connected in parallel across 220 V. Which bulb glows brighter?

Solution:

In parallel, both bulbs get the same voltage of 220 V.

The bulb with higher rated power consumes more power at the rated voltage.

Answer: The 100 W bulb glows brighter when connected in parallel.

Q4. An electric heater is rated 2 kW, 220 V. Find its resistance and current drawn from the supply.

Solution:

Given: P = 2 kW = 2000 W, V = 220 V

I = P/V = 2000/220 = 9.09 A

R = V²/P = 220²/2000 = 24.2 Ω

Answer: Current drawn is 9.09 A and resistance is 24.2 Ω.

Q5. A wire of resistance 10 Ω is cut into five equal parts. These five parts are connected in parallel. Find the equivalent resistance.

Solution:

Resistance is directly proportional to length.

Each part has resistance:

R = 10/5 = 2 Ω

Five resistors of 2 Ω each are connected in parallel:

1/R_eq = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 = 5/2

R_eq = 2/5 = 0.4 Ω

Answer: Equivalent resistance is 0.4 Ω.

Q6. A 40 W bulb is used for 5 hours daily and a 1000 W heater is used for 2 hours daily. Find total electrical energy consumed in 30 days in kWh.

Solution:

Bulb energy per day:

E_bulb = 40 W × 5 h = 200 Wh = 0.2 kWh

Heater energy per day:

E_heater = 1000 W × 2 h = 2000 Wh = 2 kWh

Total per day:

E = 0.2 + 2 = 2.2 kWh

For 30 days:

E = 2.2 × 30 = 66 kWh

Answer: Total energy consumed is 66 kWh.

Q7. A resistor of 8 Ω is connected to a battery. Heat produced in 10 seconds is 320 J. Find current and potential difference.

Solution:

Given: H = 320 J, R = 8 Ω, t = 10 s

H = I²Rt

320 = I² × 8 × 10

I² = 320/80 = 4

I = 2 A

Potential difference:

V = IR = 2 × 8 = 16 V

Answer: Current is 2 A and potential difference is 16 V.

Q8. Two resistors 3 Ω and 6 Ω are first connected in series and then in parallel across the same battery of 12 V. Find the ratio of heat produced in both combinations in the same time.

Solution:

For same voltage and same time:

H = V²t/R

Series resistance:

R_s = 3 + 6 = 9 Ω

Parallel resistance:

R_p = (3 × 6)/(3 + 6) = 18/9 = 2 Ω

Ratio of heat:

H_s : H_p = V²t/9 : V²t/2

H_s : H_p = 1/9 : 1/2 = 2 : 9

Answer: Heat produced in series and parallel combination is in the ratio 2 : 9.

Q1. Why does a thick wire have less resistance than a thin wire of the same material and length?

Answer: A thick wire has larger area of cross-section. Since resistance is inversely proportional to area, a larger area provides more path for charge flow and reduces resistance.

Q2. Why does resistance increase when length of a wire increases?

Answer: A longer wire provides a longer path for electrons. Electrons face more collisions with atoms, so resistance increases.

Q3. Why does equivalent resistance decrease in parallel combination?

Answer: In parallel combination, current gets multiple paths to flow. More paths reduce the opposition to current, so equivalent resistance decreases.

Q4. Why is a fuse wire made of a material with low melting point?

Answer: During excess current, heat produced increases. A low melting point fuse wire melts quickly and breaks the circuit, protecting appliances from damage.

Q5. Why does a 100 W bulb glow brighter than a 60 W bulb in parallel but not necessarily in series?

Answer: In parallel, both bulbs get rated voltage, so the 100 W bulb consumes more power and glows brighter. In series, same current flows through both bulbs, so the bulb with higher resistance may glow brighter.

Q6. Why does an electric heater consume more energy than an LED bulb when used for the same time?

Answer: An electric heater has much higher power rating than an LED bulb. Since energy consumed is E = Pt, higher power means more energy consumption for the same time.

Q7. Why is resistivity more useful than resistance for comparing materials?

Answer: Resistance depends on dimensions of the conductor, but resistivity depends mainly on material. Therefore, resistivity is better for comparing conducting ability of different materials.

Q8. Why is heat produced proportional to square of current?

Answer: Heat produced is given by H = I²Rt. Therefore, if current doubles, heat becomes four times. This is why excessive current is dangerous.

Q9. Why are high-power appliances not connected through very thin wires?

Answer: High-power appliances draw large current. Thin wires have higher resistance and may produce excessive heat, leading to melting, fire or energy loss.

Q10. Why does a bulb filament glow but connecting wires do not?

Answer: The filament has high resistance and becomes very hot due to Joule heating, so it glows. Connecting wires have very low resistance and do not heat up significantly.

Q1. State Ohm’s law and write its mathematical expression.

Answer: At constant temperature, current through a conductor is directly proportional to the potential difference across its ends.

V = IR

Q2. Define resistance and write its SI unit.

Answer: Resistance is the opposition offered by a conductor to the flow of current. Its SI unit is ohm, Ω.

Q3. Why is parallel arrangement used in domestic circuits?

Answer: Parallel arrangement provides same voltage to all appliances, allows independent operation and ensures that failure of one appliance does not affect others.

Q4. Write Joule’s law of heating and mention two applications.

Answer: Joule’s law of heating is:

H = I²Rt

Applications: electric heater, electric iron, electric fuse and incandescent bulb.

Q5. What is meant by 1 kWh?

Answer: One kilowatt hour is the energy consumed by an appliance of power 1 kW when used for 1 hour.

1 kWh = 3.6 × 10⁶ J

Q6. A device is rated 220 V, 100 W. What does this mean?

Answer: It means the device works properly at 220 V and consumes electrical power of 100 W when connected to a 220 V supply.

Q7. Why is tungsten used in electric bulb filament?

Answer: Tungsten has very high melting point and high resistance. It can glow at high temperature without melting easily.

Q8. Why does the cord of an electric heater not glow while the heating element glows?

Answer: The cord has low resistance, so little heat is produced. The heating element has high resistance, so large heat is produced and it glows.

Q1. Three identical resistors each of resistance R are connected first in series and then in parallel. Find the ratio of equivalent resistances.

Solution:

Series resistance:

R_s = R + R + R = 3R

Parallel resistance:

1/R_p = 1/R + 1/R + 1/R = 3/R

R_p = R/3

Ratio:

R_s : R_p = 3R : R/3 = 9 : 1

Answer: Ratio is 9 : 1.

Q2. A wire of resistance R is cut into n equal parts and all parts are connected in parallel. What is the equivalent resistance?

Solution:

Each part has resistance R/n.

When n equal resistors of resistance R/n are connected in parallel:

R_eq = (R/n)/n = R/n²

Answer: Equivalent resistance is R/n².

Q3. Two wires A and B are of the same material and length. Radius of A is twice the radius of B. Find ratio of their resistances.

Solution:

Resistance is inversely proportional to area.

A = πr²

If radius of A is 2r and radius of B is r:

Area of A : Area of B = 4 : 1

Therefore:

R_A : R_B = 1 : 4

Answer: Ratio of resistances is 1 : 4.

Q4. A 2 Ω resistor and a 4 Ω resistor are connected in parallel. This combination is connected in series with 8 Ω. If the total current is 3 A, find voltage of the battery.

Solution:

Parallel part:

R_p = (2 × 4)/(2 + 4) = 8/6 = 4/3 Ω

Total resistance:

R = 8 + 4/3 = 28/3 Ω

Battery voltage:

V = IR = 3 × 28/3 = 28 V

Answer: Battery voltage is 28 V.

Q5. A heater coil produces 800 J of heat in 10 seconds when current is 4 A. Find resistance of the coil.

Solution:

H = I²Rt

800 = 4² × R × 10

800 = 160R

R = 5 Ω

Answer: Resistance is 5 Ω.

Q6. A 220 V line supplies two appliances of power 500 W and 1100 W connected in parallel. Find total current drawn.

Solution:

Total power:

P = 500 + 1100 = 1600 W

I = P/V = 1600/220 = 7.27 A

Answer: Total current drawn is approximately 7.27 A.

Q1. Assertion: The V-I graph of an ohmic conductor is a straight line. Reason: Current is directly proportional to potential difference at constant temperature.

Answer: A. Both Assertion and Reason are true and Reason is the correct explanation.

Q2. Assertion: Equivalent resistance in parallel is less than the smallest resistance. Reason: Parallel combination provides more paths for current.

Answer: A. Both Assertion and Reason are true and Reason is the correct explanation.

Q3. Assertion: Copper is used in electrical wires. Reason: Copper has high resistivity.

Answer: C. Assertion is true but Reason is false. Copper is used because it has low resistivity.

Q4. Assertion: Heat produced in a resistor becomes four times when current is doubled. Reason: Heat produced is directly proportional to square of current.

Answer: A. Both Assertion and Reason are true and Reason is the correct explanation.

Q5. Assertion: Household appliances are connected in series. Reason: In series, each appliance gets the same voltage.

Answer: Both Assertion and Reason are false. Household appliances are connected in parallel so that each gets the same voltage.

Q6. Assertion: Electric fuse protects appliances from excess current. Reason: Fuse wire melts due to heating effect of current.

Answer: A. Both Assertion and Reason are true and Reason is the correct explanation.