About Current Electricity in FSC Physics
Current Electricity is one of the 14 main chapters in our Federal Board FSC Physics question bank (11th and 12th class), and about 8% of its real questions come from it. Tayyari Karo has 40 real questions on this chapter from 2023 to 2026; 11 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 26 medium, 13 easy, 1 hard.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 8 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
2026 Annual · 11th · Current Electricity
A potentiometer is used to compare:
- Resistance
- Current
- Potential difference
- Power
Show answer and explanation
Answer: C. Potential difference
A potentiometer compares potential differences (emfs) using a null-deflection method, without drawing current from the source being measured.
2025 1st Annual · 11th · Current Electricity
In order to achieve high accuracy, the slide wire of a potentiometer should be:
- As long as possible
- As short as possible
- Very thick
- Neither too small, nor too large
Show answer and explanation
Answer: A. As long as possible
A longer wire gives a smaller potential gradient (V per unit length), allowing the balance point to be located more precisely — increasing accuracy.
2025 2nd Annual · 11th · Current Electricity
A 10Ω resistor is connected across the terminals of a 10V battery. The power dissipation of the resistor is:
- 100 W
- 50 W
- 10 W
- 25 W
Show answer and explanation
Answer: C. 10 W
P = V²/R = 10²/10 = 10 W.
2024 Annual · 12th · Current Electricity
The best instrument for the accurate measurement of emf of a cell is:
- Potentiometer
- Ohm meter
- Voltmeter
- Ammeter
Show answer and explanation
Answer: A. Potentiometer
A potentiometer measures emf accurately because at balance it draws no current from the cell, unlike a voltmeter which always draws some current.
2024 Annual · 12th · Current Electricity
A 60 watt bulb operates at 50% efficiency; the energy dissipated (as heat) in 10 seconds is:
- 600 J
- 700 J
- 200 J
- 300 J
Show answer and explanation
Answer: D. 300 J
Total energy input = Pt = 60×10 = 600 J. At 50% efficiency, half is useful output and half (300 J) is dissipated as heat.
2023 Annual · 12th · Current Electricity
Three resistors each of 2 Ohm are connected to form a triangle. The resistance on any corner is:
- 3/4 Ohm
- 4/3 Ohm
- 4×3 Ohm
- 3×2 Ohm
Show answer and explanation
Answer: B. 4/3 Ohm
Between any two corners of the triangle, one 2Ω side is in parallel with the other two 2Ω sides in series (2+2=4Ω): R = (2×4)/(2+4) = 8/6 = 4/3 Ω.
2026 Annual · 11th · Current Electricity
Calculate the resistance of a wire 10 m long that has a diameter of 3 mm and resistivity of 2.63 × 10⁻² Ω·m.
Show model answer
R = ρL/A, where A = πd²/4
A = π(3×10⁻³)²/4 = π(9×10⁻⁶)/4 ≈ 7.07×10⁻⁶ m²
R = (2.63×10⁻²)(10)/(7.07×10⁻⁶) = 0.263/(7.07×10⁻⁶)
R ≈ 3.72 × 10⁴ Ω
2025 2nd Annual · 12th · Current Electricity
State Ohm's law. Derive its mathematical form and give its limitations.
Show model answer
Ohm's law: At constant temperature, the current I flowing through a conductor is directly proportional to the potential difference V applied across its ends: I ∝ V, so V = IR, where R (resistance) is the constant of proportionality.
Derivation: Since I ∝ V, we can write V = IR, where R = V/I is a constant for a given conductor at a given (constant) temperature.
Limitations:
1. Ohm's law holds only if the temperature (and other physical conditions) of the conductor remain constant.
2. It does not apply to non-ohmic devices, such as diodes, transistors, and thermistors, whose V-I graph is not a straight line through the origin.
3. …
The full model answer is in the app.