A Level Physics 21.1 Characteristics of Alternating Currents MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Section 21.1 of Cambridge AS & A Level Physics (9702) covers alternating currents: peak and instantaneous values, root-mean-square (rms) values, AC power in resistive loads, and reading frequency and voltage from an oscilloscope trace. This is an A Level (A2) topic, and these 30 original practice questions take you from the basics of a sinusoidal waveform right through to fully worked oscilloscope and power calculations.

Formula recap

  • Instantaneous current: I = I0 sin(2πft), where I0 is the peak current and f is the frequency
  • rms values: Irms = I0 / √2 ≈ 0.707 I0  and  Vrms = V0 / √2 ≈ 0.707 V0
  • The rms value of an AC is the steady (DC) value that produces the same heating effect in a resistor
  • Mean power in a resistive load: P = Irms Vrms = Irms2 R = Vrms2 / R
  • Peak (maximum instantaneous) power = 2 × mean power for a sinusoidal supply

For the full list of equations for this topic, see the A Level Physics Formula Sheet.

Alternating Current Basics

Q1. Which statement best describes an alternating current?

  • (A) A current that steadily increases in size
  • (B) A current that periodically reverses its direction
  • (C) A current that flows in one direction only
  • (D) A current that decreases to zero and then stays at zero
Show Answer

Answer: B – An alternating current (AC) reverses its direction periodically. For a sinusoidal supply, its magnitude also varies smoothly between zero and the peak value in each half-cycle.

Q2. A mains supply has a frequency of 50 Hz. What is the period of the supply?

  • (A) 0.0020 s
  • (B) 0.0050 s
  • (C) 0.020 s
  • (D) 50 s
Show Answer

Answer: C – Period T = 1 / f = 1 / 50 = 0.020 s, i.e. each complete cycle lasts 20 ms.

Q3. The peak value of a sinusoidal alternating current is

  • (A) the maximum value reached during each cycle
  • (B) the average value over a complete cycle
  • (C) the root-mean-square value
  • (D) the value of the current at every instant
Show Answer

Answer: A – The peak value I0 (or V0) is the maximum (amplitude) value of the current (or voltage) in each cycle. The current takes this value only instantaneously, twice per cycle.

Q4. A sinusoidal current is given by I = I0 sin(ωt) and has period T. What is the current at time t = T/4?

  • (A) zero
  • (B) 0.50 I0
  • (C) 0.71 I0
  • (D) I0
Show Answer

Answer: D – At t = T/4, ωt = (2π/T) × (T/4) = π/2, and sin(π/2) = 1, so I = I0: a quarter of a period after zero, the current is at its positive peak.

Q5. For the same current I = I0 sin(ωt), what is the current at time t = T/2?

  • (A) I0
  • (B) zero
  • (C) 0.50 I0
  • (D) −I0
Show Answer

Answer: B – At t = T/2, ωt = π and sin(π) = 0, so the current is zero. It reaches its negative peak −I0 later, at t = 3T/4.

Q6. The mean value of a sinusoidal alternating current over a complete cycle is zero, yet the current can light a lamp. This is because

  • (A) the heating effect depends on I2, which is positive in both half-cycles
  • (B) the lamp stores charge during the first half-cycle
  • (C) the mean value is not really zero
  • (D) the lamp responds only to the positive half-cycle
Show Answer

Answer: A – The power dissipated in a resistor is I2R. Squaring removes the sign, so both half-cycles heat the filament equally even though the signed currents average to zero over a full cycle.

Q7. In a simple AC generator, a flat coil rotates at a steady rate in a uniform magnetic field. The induced EMF is instantaneously greatest when

  • (A) the plane of the coil is perpendicular to the field lines
  • (B) the magnetic flux linkage of the coil is at its maximum
  • (C) the plane of the coil is parallel to the field lines
  • (D) the coil is momentarily at rest
Show Answer

Answer: C – When the coil plane is parallel to the field, the flux linkage is zero but it is changing at its greatest rate, so the induced EMF (proportional to the rate of change of flux linkage) is at its peak. When the flux linkage is maximum, its rate of change is zero and the EMF is zero.

Q8. An alternating current is described by I = 5.0 sin(100πt), with I in amperes and t in seconds. What is the frequency of the current?

  • (A) 5.0 Hz
  • (B) 25 Hz
  • (C) 100 Hz
  • (D) 50 Hz
Show Answer

Answer: D – Comparing with I = I0 sin(2πft), the angular frequency is ω = 100π = 2πf, so f = 100π / (2π) = 50 Hz. (The peak current is 5.0 A.)

Q9. An alternating current has a period of 0.040 s. Its frequency is

  • (A) 25 Hz
  • (B) 0.025 Hz
  • (C) 40 Hz
  • (D) 250 Hz
Show Answer

Answer: A – f = 1 / T = 1 / 0.040 = 25 Hz.

Q10. How many times does a 50 Hz alternating current change direction in one second?

  • (A) 50 times
  • (B) 100 times
  • (C) 25 times
  • (D) 200 times
Show Answer

Answer: B – The current reverses twice in each complete cycle (once at each zero crossing from positive to negative and back). With 50 cycles per second, there are 2 × 50 = 100 reversals per second.

Root-Mean-Square Values

Q11. The root-mean-square (rms) value of an alternating current is defined as the value of the steady (direct) current that

  • (A) has the same peak value as the AC
  • (B) transfers the same total charge in one cycle
  • (C) produces the same heating effect in a given resistor
  • (D) flows for the same total time
Show Answer

Answer: C – The rms value links AC to an equivalent DC effect: a DC current of Irms amperes dissipates exactly the same power in a resistor as the alternating current does on average.

Q12. A sinusoidal alternating current has a peak value of 4.0 A. What is its rms value?

  • (A) 2.8 A
  • (B) 4.0 A
  • (C) 5.7 A
  • (D) 2.0 A
Show Answer

Answer: A – Irms = I0 / √2 = 4.0 / 1.414 = 2.83 A ≈ 2.8 A.

Q13. An AC supply has an rms voltage of 12 V. What is its peak voltage?

  • (A) 8.5 V
  • (B) 12 V
  • (C) 24 V
  • (D) 17 V
Show Answer

Answer: D – V0 = Vrms × √2 = 12 × 1.414 = 17.0 V ≈ 17 V.

Q14. The mains voltage is quoted as 230 V. This is the rms value. What is the peak value of the mains voltage?

  • (A) 163 V
  • (B) 325 V
  • (C) 230 V
  • (D) 460 V
Show Answer

Answer: B – V0 = 230 × √2 = 230 × 1.414 = 325 V. The mains voltage actually swings between +325 V and −325 V each cycle; 230 V is the equivalent DC (heating) value.

Q15. The ordinary (plain) mean value of an alternating current over a full cycle cannot be used to calculate its heating effect because

  • (A) the positive and negative half-cycles cancel in the mean, but heating depends on I2, which is always positive
  • (B) the mean value is too difficult to measure with a meter
  • (C) the mean value changes whenever the frequency changes
  • (D) alternating currents flow too quickly to have a mean value
Show Answer

Answer: A – The signed mean is zero, which would wrongly predict no heating at all. Because power depends on I2, we square first, then average, then take the square root: this gives the rms value, I0/√2.

Q16. For a sinusoidal current I = I0 sin(ωt), the mean value of I2 averaged over a complete cycle is

  • (A) I02
  • (B) zero
  • (C) I02 / 2
  • (D) 2 I02
Show Answer

Answer: C – The mean of sin2(ωt) over a full cycle is 1/2, so the mean of I2 is I02/2. Taking the square root gives Irms = I0/√2.

Q17. A sinusoidal voltage has a peak value of 28 V. Its rms value is approximately

  • (A) 14 V
  • (B) 28 V
  • (C) 40 V
  • (D) 20 V
Show Answer

Answer: D – Vrms = V0 / √2 = 28 / 1.414 = 19.8 V ≈ 20 V.

Q18. A resistor is connected first to a 12 V DC supply and then to an AC supply. The heating effect is the same in both cases. What is the peak value of the AC voltage?

  • (A) 12 V
  • (B) 17 V
  • (C) 8.5 V
  • (D) 24 V
Show Answer

Answer: B – Equal heating means the rms value of the AC equals the DC value, so Vrms = 12 V. The peak is V0 = 12 × √2 = 17 V.

Q19. For a sinusoidal alternating current, the rms value is what fraction of the peak value?

  • (A) about 0.71
  • (B) exactly 0.50
  • (C) about 1.41
  • (D) about 0.64
Show Answer

Answer: A – Irms = I0/√2 = 0.707 I0, so the rms value is about 71% of the peak value. (The factor 1.41 is √2, used the other way: peak = 1.41 × rms.)

Q20. Two sinusoidal supplies are connected in turn to identical resistors. Supply X has a peak current of 10 A. Supply Y has an rms current of 10 A. Which supply produces the greater heating effect?

  • (A) Supply X
  • (B) Both produce the same heating
  • (C) Supply Y
  • (D) It cannot be decided without the frequency
Show Answer

Answer: C – Heating is fixed by the rms current. Supply X has Irms = 10 / √2 = 7.1 A, while supply Y has Irms = 10 A, so Y dissipates more power (in fact, twice as much, since power ∝ Irms2).


AC Power & Oscilloscope Measurements

Q21. An electric heater is connected to a 230 V rms mains supply and draws an rms current of 5.0 A. What is the mean power dissipated in the heater?

  • (A) 46 W
  • (B) 575 W
  • (C) 2300 W
  • (D) 1150 W
Show Answer

Answer: D – P = Irms Vrms = 5.0 × 230 = 1150 W. rms values can be used in the familiar power formulae exactly as DC values are.

Q22. A lamp is rated at 60 W when connected to a 230 V rms supply. What rms current does it draw?

  • (A) 0.26 A
  • (B) 3.8 A
  • (C) 0.13 A
  • (D) 0.52 A
Show Answer

Answer: A – Irms = P / Vrms = 60 / 230 = 0.26 A.

Q23. A sinusoidal supply dissipates a mean power of 500 W in a resistor. What is the peak (maximum instantaneous) power dissipated?

  • (A) 250 W
  • (B) 1000 W
  • (C) 500 W
  • (D) 707 W
Show Answer

Answer: B – The peak instantaneous power is V0 I0, while the mean power is (1/2) V0 I0. So the peak power is 2 × 500 = 1000 W. The power pulsates between zero and this peak twice per cycle.

Q24. An rms current of 2.0 A flows through a resistor of resistance 30 Ω. What is the mean power dissipated?

  • (A) 60 W
  • (B) 15 W
  • (C) 120 W
  • (D) 240 W
Show Answer

Answer: C – P = Irms2 R = (2.0)2 × 30 = 4.0 × 30 = 120 W.

Q25. An oscilloscope displays a sinusoidal voltage. The trace has a peak-to-peak height of 6.0 cm and the Y-gain is set to 2.0 V/cm. What is the peak voltage of the signal?

  • (A) 6.0 V
  • (B) 12 V
  • (C) 3.0 V
  • (D) 24 V
Show Answer

Answer: A – Peak-to-peak voltage = 6.0 cm × 2.0 V/cm = 12 V. The peak voltage is half of this: V0 = 12 / 2 = 6.0 V.

Q26. On an oscilloscope screen 10 cm wide, exactly three complete cycles of a sinusoidal signal are displayed. The time-base is set to 2.0 ms/cm. What is the frequency of the signal?

  • (A) 15 Hz
  • (B) 67 Hz
  • (C) 300 Hz
  • (D) 150 Hz
Show Answer

Answer: D – Three cycles occupy 10 cm, so one cycle occupies 10 / 3 = 3.33 cm. Period T = 3.33 cm × 2.0 ms/cm = 6.67 ms = 6.67 × 10−3 s. Frequency f = 1 / T = 1 / (6.67 × 10−3) = 150 Hz.

Q27. An oscilloscope trace shows one complete cycle occupying 5.0 cm horizontally, with the time-base set to 1.0 ms/cm. The frequency of the signal is

  • (A) 20 Hz
  • (B) 200 Hz
  • (C) 50 Hz
  • (D) 500 Hz
Show Answer

Answer: B – Period T = 5.0 cm × 1.0 ms/cm = 5.0 ms = 5.0 × 10−3 s, so f = 1 / T = 200 Hz.

Q28. A sinusoidal signal produces an oscilloscope trace whose peak height (from the centre line to the top of the trace) is 4.0 cm. The Y-gain is 5.0 V/cm. What is the rms voltage of the signal?

  • (A) 20 V
  • (B) 28 V
  • (C) 14 V
  • (D) 10 V
Show Answer

Answer: C – Peak voltage V0 = 4.0 cm × 5.0 V/cm = 20 V. Then Vrms = V0 / √2 = 20 / 1.414 = 14.1 V ≈ 14 V.

Q29. A sinusoidal voltage with peak value 100 V is applied across a resistor of resistance 25 Ω. What is the mean power dissipated in the resistor?

  • (A) 400 W
  • (B) 200 W
  • (C) 100 W
  • (D) 800 W
Show Answer

Answer: B – Mean power = V02 / (2R) = (100)2 / (2 × 25) = 10 000 / 50 = 200 W. (400 W is the peak instantaneous power, V02/R; the mean is half of this.)

Q30. An electric heater is rated at 2.0 kW for use on a 230 V rms supply. What is the resistance of its heating element?

  • (A) 8.7 Ω
  • (B) 0.115 Ω
  • (C) 53 Ω
  • (D) 26 Ω
Show Answer

Answer: D – P = Vrms2 / R, so R = Vrms2 / P = (230)2 / 2000 = 52 900 / 2000 = 26.45 Ω ≈ 26 Ω.

Keep practising

Browse the full range of section-by-section questions on the Physics MCQs page, and keep the key equations for every topic in one place with the A Level Physics Formula Sheet.

These are original practice questions written for QuaidianPoint, based on the topics in the Cambridge 9702 syllabus. QuaidianPoint is not affiliated with Cambridge Assessment International Education, and these questions are not Cambridge past-paper questions.