A Level Physics (A2) chapter 10: Alternating currents MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
47 exam-style questions on Alternating currents, 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.
Characteristics of alternating currents24 questions
Rectification and smoothing23 questions
Ten Alternating currents questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A sinusoidal alternating current of peak value 3.0 A passes through a 20 Ω resistor. What is the mean power dissipated in the resistor?
A
180 W
B
90 W
C
130 W
D
60 W
Show answer
Answer: B. The maximum power is I02R = 3.02 × 20 = 180 W. For a sinusoidal current the mean power is half the maximum: 90 W. This is the same as Ir.m.s.2R with Ir.m.s. = 3.0/√2 A.
2. An alternating p.d. has a square waveform: it is +10 V for half of each cycle and −10 V for the other half, switching instantly between the two. What is its r.m.s. value?
A
0 V
B
5.0 V
C
7.1 V
D
10 V
Show answer
Answer: D. V2 is 100 V2 at every instant, so the mean of V2 is 100 V2 and Vr.m.s. = 10 V. The factor 1/√2 applies only to sinusoidal waveforms, and 0 V is the mean, not the r.m.s. value.
3. An alternating current has a period of 20 ms. What is its frequency?
A
0.050 Hz
B
20 Hz
C
314 Hz
D
50 Hz
Show answer
Answer: D. f = 1/T = 1/0.020 s = 50 Hz. Forgetting to convert milliseconds gives 0.050 Hz; 314 is the angular frequency 2π/T in rad s−1.
4. A resistor dissipates power P when connected to a 12 V d.c. supply. It is then connected to a sinusoidal a.c. supply of peak value 12 V. What is the mean power dissipated now?
A
P/2
B
P/√2
C
P
D
P/4
Show answer
Answer: A. With the a.c. supply the maximum power, at the peaks, is 122/R = P. The mean power for a sinusoidal supply is half the maximum, P/2. Dividing by √2 applies to the p.d., not to the power.
5. One of the four diodes in a bridge rectifier breaks and becomes an open circuit. The rest of the circuit is unchanged. What is the output across the load now?
A
full-wave rectified, with half the peak value
B
zero at all times
C
half-wave rectified
D
an unrectified alternating p.d.
Show answer
Answer: C. Each half-cycle uses a different pair of diodes in series. With one diode open, its pair cannot conduct, so one half-cycle is lost; the other pair still passes the other half-cycle.
6. A smoothed full-wave rectifier gives a ripple of 0.40 V. The time constant of the capacitor and load is much greater than the period of the supply. One diode of the bridge breaks and becomes an open circuit, so the circuit now gives half-wave rectification. Nothing else changes. What is the approximate ripple now?
A
0.20 V
B
0.40 V
C
0.57 V
D
0.80 V
Show answer
Answer: D. When RC is much greater than the discharge time, the fall in p.d. is roughly proportional to the time between peaks. Half-wave rectification doubles that time (from T/2 to T), so the ripple roughly doubles to 0.80 V.
7. A graph of the output p.d. of a rectifier against time shows humps, all on the same side of the time axis. Each hump lasts 10 ms and is followed by a gap of 10 ms in which the p.d. is zero. What type of rectification is this, and what is the supply frequency?
A
half-wave, 50 Hz
B
half-wave, 100 Hz
C
full-wave, 50 Hz
D
full-wave, 100 Hz
Show answer
Answer: A. The gaps show that every other half-cycle is removed, so this is half-wave rectification. One hump plus one gap is one cycle of 20 ms, so f = 1/0.020 = 50 Hz. Using only the 10 ms hump gives 100 Hz.
8. A bridge rectifier connects a load resistor to a sinusoidal supply. Which statement about the currents is correct?
A
The current in the load is always in the same direction, but the current in the supply reverses every half-cycle.
B
The current in the supply is always in one direction, but the current in the load reverses every half-cycle.
C
The currents in the supply and in the load both reverse every half-cycle.
D
The currents in the supply and in the load are both always in the same direction.
Show answer
Answer: A. The supply is still alternating, so its current reverses. The bridge steers the current so that it always enters the load at the same end, giving a unidirectional (full-wave rectified) load current.
9. Which statement correctly describes the peak value of an alternating current?
A
It is the greatest magnitude of the current measured from zero.
B
It is the difference between the greatest positive and greatest negative currents.
C
It is the steady current that gives the same mean power in a resistor.
D
It is the average magnitude of the current over half a cycle.
Show answer
Answer: A. The peak value (amplitude) is the maximum magnitude of the current measured from zero. The second option is the peak-to-peak value and the third describes the r.m.s. value.
10. A sinusoidal p.d. is half-wave rectified using an ideal diode. For what fraction of each cycle of the supply is the output p.d. zero?
A
none of the cycle
B
one quarter of the cycle
C
one half of the cycle
D
three quarters of the cycle
Show answer
Answer: C. The diode conducts for one half-cycle and blocks the other, so the output is zero for half of each cycle. With full-wave rectification it is zero only at instants.
47, each with the answer and a short explanation, across 2 topics: Characteristics of alternating currents, Rectification and smoothing. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Physics (A2) chapter 10 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.
