MDCAT Physics chapter 9: Physics · Current electricity MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
18 exam-style questions on Physics · Current electricity, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Current electricity questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A uniform wire of resistance 12 Ω is cut into four equal pieces, and the pieces are connected in parallel. The resistance of the combination is
A
3 Ω
B
0.75 Ω
C
48 Ω
D
1.5 Ω
Show answer
Answer: B. Each piece has a resistance of 12/4 = 3 Ω, and four 3 Ω resistors in parallel give 3/4 = 0.75 Ω. 3 Ω is a single piece, and 48 Ω multiplies instead of dividing.
2. A lamp is rated 100 W at 220 V. If it is connected to a 110 V supply and its resistance stays the same, its power is
A
50 W
B
25 W
C
100 W
D
12.5 W
Show answer
Answer: B. P = V2/R with R constant, so halving V cuts the power to one-quarter: 100/4 = 25 W. 50 W wrongly takes the power as proportional to V.
3. A cell drives a current of 0.50 A through a 2.0 Ω resistor and a current of 0.25 A through a 5.0 Ω resistor. Its EMF and internal resistance are
A
1.25 V and 0.50 Ω
B
2.0 V and 2.0 Ω
C
1.5 V and 1.0 Ω
D
1.5 V and 0.50 Ω
Show answer
Answer: C. E = I(R + r) in both cases: 0.50(2.0 + r) = 0.25(5.0 + r), so 1.0 + 0.50r = 1.25 + 0.25r, giving r = 1.0 Ω and E = 0.50 × 3.0 = 1.5 V. Each of the other pairs fits only one of the two readings, or neither.
4. Two wires A and B are made of the same metal and have the same length, but the diameter of A is twice that of B. The ratio of their resistances RA : RB is
A
1 : 2
B
1 : 4
C
4 : 1
D
2 : 1
Show answer
Answer: B. R = ρL/A and the area is proportional to the square of the diameter, so A has 4 times the area and one-quarter of the resistance: 1 : 4. 1 : 2 forgets to square the diameter.
5. A Wheatstone bridge is balanced, so the galvanometer connected across it shows no deflection. This is because
A
no current is drawn from the battery at balance
B
all four arms of the bridge carry equal currents
C
the galvanometer has a much higher resistance than the arms
D
the two ends of the galvanometer are at the same potential
Show answer
Answer: D. At balance the two points joined by the galvanometer are at the same potential, so no current flows through it; then P/Q = R/S. Current still flows from the battery through the arms, and the arm currents need not be equal.
6. A Wheatstone bridge with arms P = 2 Ω, Q = 3 Ω, R = 4 Ω and S = 6 Ω is balanced (P/Q = R/S). A 6 Ω resistor is now connected in parallel with S. To restore balance, R must be changed to
A
8 Ω
B
2 Ω
C
4 Ω
D
3 Ω
Show answer
Answer: B. S and the extra 6 Ω in parallel give 3 Ω. Balance then needs R = P × 3/Q = 2 × 3/3 = 2 Ω. Adding the two 6 Ω resistors in series (12 Ω) instead would give 8 Ω.
7. A 4 Ω resistor is connected in series with a parallel pair of 6 Ω and 12 Ω resistors across a 12 V battery of negligible internal resistance. The current in the 6 Ω resistor is
A
1.5 A
B
0.5 A
C
2.0 A
D
1.0 A
Show answer
Answer: D. 6 Ω and 12 Ω in parallel give 4 Ω, so the total is 8 Ω and the battery current is 12/8 = 1.5 A. The parallel pair has 1.5 × 4 = 6 V across it, so the 6 Ω resistor carries 6/6 = 1.0 A and the 12 Ω carries 0.5 A.
8. Two resistors have a combined resistance of 9 Ω when connected in series and 2 Ω when connected in parallel. Their values are
A
4 Ω and 5 Ω
B
1 Ω and 8 Ω
C
3 Ω and 6 Ω
D
2 Ω and 7 Ω
Show answer
Answer: C. R1 + R2 = 9 and R1R2/(R1 + R2) = 2, so R1R2 = 18: the pair is 3 Ω and 6 Ω. Every pair offered adds up to 9 Ω, but 4 Ω and 5 Ω, for example, give about 2.2 Ω in parallel.
9. A steady current of 1.6 A flows in a wire. Taking e = 1.6 × 10−19 C, the number of electrons that pass any cross-section of the wire each second is
A
1.0 × 1019
B
2.6 × 10−19
C
1.0 × 10−19
D
1.6 × 1019
Show answer
Answer: A. I = Q/t = ne/t, so the number per second is I/e = 1.6 ÷ (1.6 × 10−19) = 1.0 × 1019. Multiplying I by e instead of dividing gives 2.6 × 10−19.
10. The potential difference across the terminals of a battery is 12 V when no current is drawn and falls to 10 V when the battery delivers a current of 4.0 A. Its internal resistance is
A
2.5 Ω
B
0.50 Ω
C
3.0 Ω
D
2.0 Ω
Show answer
Answer: B. The open-circuit potential difference is the EMF, 12 V. The lost volts are 12 − 10 = 2.0 V = Ir, so r = 2.0/4.0 = 0.50 Ω. 2.5 Ω (10/4.0) is the external resistance.
Physics · Current electricity: questions students ask
How many Physics · Current electricity MCQs are there?
18, each with the answer and a short explanation, across 2 topics: Resistance and Ohm's law, Circuits, EMF and power. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 9 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
