Extra long-answer questions (7 marks each) for the chapter:11 "Electricity" in Class 10 CBSE Science, including numerical.
1. Explain Ohm’s Law with a labelled circuit diagram. Derive its mathematical expression.
Answer:
Ohm’s Law states that the
current (I) is directly proportional to the voltage (V)
applied across a conductor, provided the temperature remains constant.
Formula:
V=IRV = IRV=IR.
Derivation:
Given in the previous answer.
Diagram:
(Show a circuit with a battery, resistor, and ammeter).
2. Compare series and parallel circuits. Which one is better for household connections?
Answer:
|
Factor |
Series Circuit |
Parallel Circuit |
|---|---|---|
|
Current |
Same for all resistors |
Different in each branch |
|
Voltage |
Divided across resistors |
Same across all branches |
|
Resistance |
Higher |
Lower |
|
Failure Effect |
If one component fails, the circuit breaks |
Other appliances keep working |
|
Application |
Decorative lights |
Household wiring |
Conclusion: Parallel circuits are used in homes because each appliance receives the same voltage.
3. State Joule’s Law of Heating and derive its equation. Give two applications.
Answer:
Joule’s Law:
Heat produced in a conductor is
H=I2RtH = I^2 R tH=I2Rt
Derivation:
Already given.
Applications:
Electric
iron, heater, toaster
(converts electricity into heat).
Electric
fuse
(melts when excess current flows).
4. What is electrical power? Derive its formulas and explain the commercial unit of energy.
Answer:
Power formula:
P=VI,P=I2R,P=V2RP = VI, \quad P = I^2 R, \quad P = \frac{V^2}{R}P=VI,P=I2R,P=RV2
Commercial Unit:
1 kilowatt-hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J
.
Example: A
100W bulb used for 10 hours
consumes
E=100×10=1000Wh=1kWhE = 100 \times 10 = 1000Wh = 1kWhE=100×10=1000Wh=1kWh
5. Why is electrical energy transmitted at high voltage over long distances?
Answer:
Power loss in transmission lines:
Ploss=I2RP_{\text{loss}} = I^2 RPloss=I2R
To reduce
current (I)
, voltage is
increased
using transformers.
Advantage:
Less power loss, efficient transmission.
6. Why is a fuse used in electrical circuits? Explain the working of a fuse with an example.
Answer:
A fuse
protects appliances from overloading
.
Material:
Low melting point (tin-lead alloy).
Working:
If current exceeds the limit, the fuse
melts
and breaks the circuit.
Example:
5A fuse in a 1000W, 220V appliance:
I=PV=1000220≈4.5AI = \frac{P}{V} = \frac{1000}{220} \approx 4.5AI=VP=2201000≈4.5A
A 5A fuse will work safely.
7. How does resistance depend on length, cross-sectional area, and material?
Answer:
Formula:
R=ρLAR = \rho \frac{L}{A}R=ρAL
Where ρ\rhoρ is resistivity, LLL is length, and AAA is area.
More length = More resistance
.
More area = Less resistance
.
Different materials have different resistivity
.
8. Explain how a household circuit works with a labelled diagram.
Answer:
Parallel wiring
ensures all devices get the same voltage.
Includes
fuse, switches, and earthing
.
Diagram:
(Main supply → Fuse → Meter → Distribution box → Appliances).
9. A 10Ω resistor is connected to a 12V battery. Find the current and power.
Solution:
I=VR=1210=1.2AI = \frac{V}{R} = \frac{12}{10} = 1.2AI=RV=1012=1.2A P=VI=12×1.2=14.4WP = VI = 12 \times 1.2 = 14.4WP=VI=12×1.2=14.4W
10. Find the total resistance when three resistors of 5Ω, 10Ω, and 15Ω are connected in series.
Solution:
Rtotal=5+10+15=30ΩR_{\text{total}} = 5 + 10 + 15 = 30ΩRtotal=5+10+15=30Ω
11. A heater draws 10A from a 220V supply. Calculate its power and energy consumed in 2 hours.
Solution:
P=VI=220×10=2200W=2.2kWP = VI = 220 \times 10 = 2200W = 2.2kWP=VI=220×10=2200W=2.2kW E=P×t=2.2×2=4.4kWhE = P \times t = 2.2 \times 2 = 4.4kWhE=P×t=2.2×2=4.4kWh
A wire has resistance
20Ω
. If its length is doubled and area is halved, find the new resistance. (
Ans:
80Ω
)
A
100W, 220V
bulb is used for
10 hours
. Find energy consumed. (
Ans:
1kWh
)
Find the resistance of a
2m
long copper wire of cross-section
1mm²
, resistivity ρ=1.7×10−8Ωm\rho = 1.7 × 10
^{
-8} \Omega
mρ
=1.7×10−8Ωm. (
Ans:
0.034Ω
)
A
3Ω, 6Ω, and 9Ω
resistors are connected in
parallel
. Find total resistance. (
Ans:
1.64Ω
)
Find the current in a
40W, 220V
lamp. (
Ans:
0.18A
)
12. A heater coil is made of a wire having a resistance of 100Ω. It operates on a 220V mains supply. Calculate:
(a) The current flowing through it.
(b) The power consumed.
(c) The heat energy produced in 5 minutes.
Solution:
Given:
R=100ΩR = 100Ω
R
=
100Ω
,
V=220VV = 220V
V
=
220V
,
t=5 min=5×60=300st = 5 \
text{ min
} = 5 \times 60 = 300s
t
=
5 min
=
5
×
60
=
300s
(a) Current Flowing
Using Ohm’s Law:
I=VR=220100=2.2AI = \frac{V}{R} = \frac{220}{100} = 2.2AI=RV=100220=2.2A
(b) Power Consumed
Using Power formula:
P=VI=220×2.2=484WP = VI = 220 \times 2.2 = 484WP=VI=220×2.2=484W
(c) Heat Energy Produced
Using Joule’s law of heating:
H=P×t=484×300=1,45,200J=145.2kJH = P \times t = 484 \times 300 = 1,45,200J = 145.2kJH=P×t=484×300=1,45,200J=145.2kJ
13. A 60W electric bulb is used for 6 hours daily. Find:
(a) The total energy consumed in 1 month.
(b) The cost of electricity if the rate is ₹7 per unit.
Solution:
Given:
P=60W=0.06kWP = 60W = 0.06kW
P
=
60W
=
0.06kW
,
t=6 hours/
dayt
= 6 \
text{ hours
/day}
t
=
6 hours/day
,
days=30\text{days} = 30
days
=
30
,
Rate = ₹7 per unit
(a) Total Energy Consumed
E=P×t×days=0.06×6×30=10.8kWhE = P \times t \times \text{days} = 0.06 \times 6 \times 30 = 10.8kWhE=P×t×days=0.06×6×30=10.8kWh
(b) Cost of Electricity
Cost=10.8×7=₹75.6\text{Cost} = 10.8 \times 7 = ₹75.6Cost=10.8×7=₹75.6
14. A 500W electric heater is used for 5 hours daily. Find:
(a) The total energy consumed in 20 days.
(b) The total cost if the rate is ₹6 per unit.
Solution:
Given:
P=500W=0.5kWP = 500W = 0.5kW
P
=
500W
=
0.5kW
,
t=5 hours/
dayt
= 5 \
text{ hours
/day}
t
=
5 hours/day
,
total days = 20
Rate = ₹6 per unit
(a) Total Energy Consumed
E=P×t×days=0.5×5×20=50kWhE = P \times t \times \text{days} = 0.5 \times 5 \times 20 = 50kWhE=P×t×days=0.5×5×20=50kWh
(b) Cost of Electricity
Cost=50×6=₹300\text{Cost} = 50 \times 6 = ₹300Cost=50×6=₹300
15. An electric kettle has a resistance of 20Ω and operates on a 220V supply. Find:
(a) The current through the kettle.
(b) The power consumed.
(c) The energy used in 10 minutes.
Solution:
Given:
R=20ΩR = 20Ω
R
=
20Ω
,
V=220VV = 220V
V
=
220V
,
t=10 min=600st = 10 \
text{ min
} = 600s
t
=
10 min
=
600s
(a) Current Flowing
Using Ohm’s Law:
I=VR=22020=11AI = \frac{V}{R} = \frac{220}{20} = 11AI=RV=20220=11A
(b) Power Consumed
Using Power formula:
P=VI=220×11=2420W=2.42kWP = VI = 220 \times 11 = 2420W = 2.42kWP=VI=220×11=2420W=2.42kW
(c) Energy Consumed
E=P×t=2.42×6003600=0.4kWhE = P \times t = 2.42 \times \frac{600}{3600} = 0.4kWhE=P×t=2.42×3600600=0.4kWh
