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CodeVID-P10-03-CH-01
Electricity — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The SI unit of potential difference is the:
- A.ampere
- B.volt
- C.ohm
- D.watt
2.
Ohm's law is:
- A.$V=\frac{I}{R}$
- B.$V=IR$
- C.$I=VR$
- D.$R=VI$
3.
The SI unit of resistivity is:
- A.$\Omega$
- B.$\Omega\ \text{m}$
- C.$\Omega/\text{m}$
- D.$\text{V/A}$
4.
Two $4\ \Omega$ resistors in parallel give:
- A.$8\ \Omega$
- B.$4\ \Omega$
- C.$2\ \Omega$
- D.$1\ \Omega$
5.
Joule's law of heating is $H=$
- A.$IRt$
- B.$I^2Rt$
- C.$VRt$
- D.$\frac{V}{R}t$
6.
$1\ \text{kWh}$ equals:
- A.$3.6\times10^{3}\ \text{J}$
- B.$3.6\times10^{6}\ \text{J}$
- C.$3600\ \text{J}$
- D.$1000\ \text{J}$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
A charge of 24 C flows in 6 s. Find the current.
8.
Find the equivalent resistance of $3\ \Omega$ and $6\ \Omega$ in parallel.
9.
State Joule's law of heating.
10.
Why are appliances connected in parallel at home?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
A 6 V battery drives 0.5 A through a resistor. Find $R$ and the power dissipated.
12.
A $10\ \Omega$ resistor carries 2 A for 5 minutes. Find the heat produced.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
A $6\ \Omega$ and $3\ \Omega$ resistor are in parallel; this is in series with $4\ \Omega$ across a 12 V battery. Find the total resistance, the battery current and the power supplied.
14.
Derive $P=VI=I^2R=\frac{V^2}{R}$ and find the resistance and current of a 100 W, 200 V bulb.
Answer Key
Section A — Multiple Choice Questions
- (B) volt
- (B) $V=IR$
- (B) $\Omega\ \text{m}$
- (C) $2\ \Omega$
- (B) $I^2Rt$
- (B) $3.6\times10^{6}\ \text{J}$
Section B — Short Answer (2 marks)
- $4\ \text{A}$.
- $2\ \Omega$.
- $H=I^2Rt$.
- Each gets full voltage and can be switched independently.
Section C — Short Answer (3 marks)
- $R=12\ \Omega$; $P=VI=3\ \text{W}$.
- $H=4\times10\times300=12000\ \text{J}$.
Section D — Long Answer (5 marks)
- Parallel $=2\ \Omega$; total $=6\ \Omega$; current $=2\ \text{A}$; power $=VI=24\ \text{W}$.
- Using $P=VI$ and $V=IR$; $R=\frac{V^2}{P}=400\ \Omega$, $I=\frac{P}{V}=0.5\ \text{A}$.
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