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Class 12 Current Electricity worksheet
CBSE Physics. Resistivity, cells, Kirchhoff's rules, Wheatstone bridge and metre bridge, power. Every sheet has 20 questions in three sections and an answer key, and the numbers change every time.
Sample questions
Real questions from this chapter, as the app sets them. Tap Answer to see the answer and working.
Section A · multiple choice
- 1.
A current of 4 A flows for 21 s. The charge that passes is:
- 84 C
- 25 C
- 5.25 C
- 0.19 C
Answer
(a) 84 C
Q = It
- 2.
A wire 18 m long with cross-section 8 mm² has resistivity 37 × 10⁻⁸ Ω m. Its resistance is:
- 83.25 Ω
- 0.08325 Ω
- 0.8325 Ω
- 0.16 Ω
Answer
(c) 0.8325 Ω
R = ρL/A
- 3.
The equivalent resistance of 2 Ω, 5 Ω in parallel is:
- 2.86 Ω
- 7 Ω
- 1.43 Ω
- 2.43 Ω
Answer
(c) 1.43 Ω
1/R = Σ 1/Rᵢ
Section B · short answer
- 4.
A 7 Ω resistor is in series with 2 Ω and 7 Ω in parallel, across an ideal 20 V battery. Find the current through each resistor.
Answer
7 Ω: 2.34 A; 2 Ω: 1.82 A; 7 Ω: 0.52 A
R = 7 + 1.56 = 8.56 Ω
- 5.
6 identical cells, each of emf 4 V and internal resistance 1 Ω, drive a 7 Ω resistor. Find the current when the cells are joined (a) in series, (b) in parallel.
Answer
(a) 1.85 A; (b) 0.56 A
series: nE/(R + nr); parallel: E/(R + r/n)
Section C · long answer
- 6.
Cells of 11 V and 8 V (no internal resistance) are joined in parallel through resistors of 4 Ω and 2 Ω respectively, and the pair drives a 4 Ω load. Using Kirchhoff's rules, find the current through each cell and through the load.
Answer
11 V cell: 1.0625 A; 8 V cell: 0.625 A (a negative value means it is being charged); load: 1.6875 A
junction potential V = 6.75 V from (11 − V)/4 + (8 − V)/2 = V/4
Fresh sheets for your child
In the app, every Current Electricity sheet is new: the app remembers the questions your child has already had and leaves them out. Print the sheet as a PDF, or use notebook mode and write the answers in a notebook.
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