Alternating currents: A Level Physics questions with answers
Free Alternating currents questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.
An alternating voltage V = 340 sin(120πt) volts is applied across a resistor of resistance 50 Ω.
(a)
State the frequency and the r.m.s. voltage.
[2]
(b)
Calculate the mean power dissipated, and the maximum power.
[2]
(c)
The supply is connected through a single diode. Calculate the mean power now dissipated in the resistor, and explain why a capacitor connected in parallel with the resistor smooths the output.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
60 Hz, 240 V
60 Hz A1 240 A1
2
(b)
1160 W, 2310 W
mean = Vrms2/R A1 maximum = 2 × mean A1
2
(c)
578 W; capacitor charges at peaks and discharges through the resistor between peaks
half-wave: power halves C1 578 A1 capacitor discharges through R when supply voltage falls B1
3
Question 2Structured · 7 marksA* challenge
A full-wave rectified 50 Hz supply has a peak output of 10 V. A 100 μF smoothing capacitor is connected across a load of 1 kΩ.
(a)
Describe the output of a full-wave rectifier without smoothing.
[2]
(b)
Calculate the time constant of the smoothing circuit.
[1]
(c)
The capacitor discharges for approximately 10 ms between peaks. Estimate the ripple voltage.
[3]
(d)
State the effect of increasing the capacitance on the ripple.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
all half-cycles positive
current in the load is always in the same direction B1 output has two positive humps per cycle (100 humps per second) B1
2
(b)
0.100 s
τ = RC = 0.100 s B1
1
(c)
0.952 V
V = 10e−0.010/RCC1 = 9.05 V C1 ripple = 0.952 V A1
3
(d)
ripple decreases
ripple decreases (larger time constant) B1
1
Question 3Structured · 4 marks
A bridge rectifier with a smoothing capacitor converts a.c. to d.c.
(a)
Explain how a single diode gives half-wave rectification.
[1]
(b)
Explain how the smoothing capacitor reduces the ripple.
[2]
(c)
State how the ripple can be reduced further.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
conducts only in one direction
the diode only conducts in one direction, so only half of each cycle passes B1
1
(b)
charges at peaks, discharges slowly through load
capacitor charges at the peaks B1 it discharges slowly through the load between peaks, maintaining the p.d. B1
2
(c)
larger capacitance or larger load resistance
use a larger capacitance (larger time constant) B1
1
Question 4Structured · 11 marks
An alternating p.d. V = 325 sin(100πt) volts is applied across a resistor of resistance 100 Ω.
(a)
Define the root-mean-square value of an alternating current.
[2]
(b)
State the frequency and the peak value of the p.d.
[2]
(c)
Calculate the r.m.s. current in the resistor.
[2]
(d)
Calculate the mean power dissipated in the resistor.
[2]
(e)
The supply is connected through a single diode. Describe the output and state how a capacitor would smooth it.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
equivalent steady current producing same mean power in a resistor
the value of the steady (direct) current B1 that produces the same mean power in a resistor as the alternating current B1
2
(b)
50 Hz; 325 V
frequency 50 Hz B1 peak 325 V B1
2
(c)
2.30 A
Vrms = 325/√2 = 230 V C1 Irms = 2.30 A A1
2
(d)
528 W
P = V02/(2R) C1 528 W A1
2
(e)
half-wave rectified (only positive halves); capacitor in parallel with load discharges between peaks; larger RC gives smoother output
half-wave rectification: current only in one half of each cycle B1 a capacitor in parallel with the load charges at each peak and discharges slowly between peaks B1 a larger RC gives less ripple B1
3
Question 5Structured · 9 marks
A sinusoidal alternating p.d. of frequency 50 Hz is displayed on an oscilloscope (Fig. 5.1). The Y-gain is set to 5 V per division.
Fig. 5.1
(a)
Use Fig. 5.1 to determine the peak value of the p.d.
[1]
(b)
Write an equation for the p.d. as a function of time.
[1]
(c)
Calculate the r.m.s. p.d.
[1]
(d)
Use Fig. 5.1 to determine the time-base setting, in ms per division.
[2]
(e)
Explain why the mains supply in a house is quoted as an r.m.s. value.
[2]
(f)
State the mean power in a 10 Ω resistor connected to this supply.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
15 V
3 divisions × 5 V = 15 V B1
1
(b)
V = 15 sin(100πt)
V = 15 sin(100πt) (or sin 314t) B1
1
(c)
10.6 V
15/√2 = 10.6 V B1
1
(d)
5.0 ms per division
period = 1/50 = 20 ms C1 one cycle is 4 divisions, so 5.0 ms per division A1
2
(e)
gives same heating effect as d.c. of same value; directly compares power
the r.m.s. value is the d.c. value that gives the same power dissipation in a resistor B1 so power calculations use P = VrmsIrmsB1
2
(f)
11.3 W
P = Vrms2/R C1 11.3 W A1
2
Question 6Structured · 10 marks
An ideal transformer has 2000 turns on its primary coil and 50 turns on its secondary coil. The primary is connected to the 230 V r.m.s. mains.
(a)
Explain how the transformer produces an alternating p.d. in the secondary coil.
[3]
(b)
Calculate the r.m.s. output p.d.
[1]
(c)
A 20 Ω resistor is connected to the secondary. Use your answer to (b) to calculate the peak current in the resistor.
[2]
(d)
Calculate the r.m.s. current in the primary coil.
[2]
(e)
State two causes of energy loss in a real transformer and how each is reduced.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
a.c. in primary; changing flux in core; links secondary, induces e.m.f. (Faraday)
alternating current in the primary produces a changing magnetic flux in the core B1 the changing flux links the secondary coil B1 an e.m.f. is induced (Faraday’s law) B1
3
(b)
5.75 V
Vs = VpNs/Np = 5.75 V B1
1
(c)
0.407 A
Irms = 0.287 A C1 peak = √2 × Irms = 0.407 A A1
2
(d)
7.19 × 10−3 A
power in = power out = 1.65 W C1 Ip = 7.19 × 10−3 A A1
2
(e)
eddy currents: laminated core; resistance of coils: thick copper wire; hysteresis: soft iron
eddy currents in the core, reduced by laminating it B1 heating in the coils, reduced by using thick low-resistance wire (or hysteresis loss, reduced by soft iron) B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in