Cambridge International A-Level Physics · topic 21
Alternating currents
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
An rms reading and a smoothing capacitor
An ideal sinusoidal AC source has peak voltage 12.0 V and frequency 50 Hz. It drives a 6.0 Ω resistor. The supplied graph gives selected voltage values over a full cycle, joined as a sinusoid. In a separate circuit an ideal full-wave rectifier charges a capacitor that supplies an approximately constant 0.020 A load; use ΔV≈IΔt/C between charging peaks. Supplied data (Time / ms; Voltage / V): (0, 0), (0.625, 2.3411), (1.25, 4.5922), (1.875, 6.6668), (2.5, 8.4853), (3.125, 9.9776), (3.75, 11.0866), (4.375, 11.7694), (5, 12), (5.625, 11.7694), (6.25, 11.0866), (6.875, 9.9776), (7.5, 8.4853), (8.125, 6.6668), (8.75, 4.5922), (9.375, 2.3411), (10, 0), (10.625, -2.3411), (11.25, -4.5922), (11.875, -6.6668), (12.5, -8.4853), (13.125, -9.9776), (13.75, -11.0866), (14.375, -11.7694), (15, -12), (15.625, -11.7694), (16.25, -11.0866), (16.875, -9.9776), (17.5, -8.4853), (18.125, -6.6668), (18.75, -4.5922), (19.375, -2.3411), (20, 0). (a) Find rms voltage, rms current and average resistor power. Explain why the zero mean voltage does not imply zero power. [3 marks] (b) Find minimum smoothing capacitance for a 0.50 V approximate ripple, and explain the effect of half-wave rectification at the same load. [3 marks]
Source waveform
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Voltage / V
Horizontal axis (x): Time / ms
Read the exact data as a table
| Time / ms | Voltage / V |
|---|---|
| 0 | 0 |
| 0.625 | 2.3411 |
| 1.25 | 4.5922 |
| 1.875 | 6.6668 |
| 2.5 | 8.4853 |
| 3.125 | 9.9776 |
| 3.75 | 11.0866 |
| 4.375 | 11.7694 |
| 5 | 12 |
| 5.625 | 11.7694 |
| 6.25 | 11.0866 |
| 6.875 | 9.9776 |
| 7.5 | 8.4853 |
| 8.125 | 6.6668 |
| 8.75 | 4.5922 |
| 9.375 | 2.3411 |
| 10 | 0 |
| 10.625 | -2.3411 |
| 11.25 | -4.5922 |
| 11.875 | -6.6668 |
| 12.5 | -8.4853 |
| 13.125 | -9.9776 |
| 13.75 | -11.0866 |
| 14.375 | -11.7694 |
| 15 | -12 |
| 15.625 | -11.7694 |
| 16.25 | -11.0866 |
| 16.875 | -9.9776 |
| 17.5 | -8.4853 |
| 18.125 | -6.6668 |
| 18.75 | -4.5922 |
| 19.375 | -2.3411 |
| 20 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find rms voltage, rms current and average resistor power. Explain why the zero mean voltage does not imply zero power.
Reveal the worked solution
- For a sinusoid, Vrms = Vpeak/√2 = 12/√2≈8.49 V. Irms = Vrms/R≈1.41 A.
- Mean heating power is Vrms²/R = (12²/2)/6 = 12 W. Using peak voltage directly would double this average.
- Voltage and current reverse together, so instantaneous resistor power v²/R stays nonnegative even though the full-cycle mean voltage is zero.
- A full-wave rectifier provides two charging peaks per 50 Hz cycle, so the time between peaks is approximately 0.010 s.
- C≈IΔt/ΔV = 0.020(0.010)/0.50 = 4.0×10⁻⁴ F = 400 μF.
- Half-wave rectification gives one charging peak per cycle, doubling Δt to 0.020 s and requiring about 800 μF. Actual ripple depends on diode drops, conduction intervals and load variation.
Answer (a) Vrms≈8.49 V; Irms≈1.41 A; mean power=12.0 W. (b) Full-wave C≈400 μF; half-wave needs≈800 μF for the same approximate ripple.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Reducing grid losses while respecting transformer limits
An ideal transformer raises an AC supply from 250 V to 10.0 kV. It delivers 20.0 kW to a transmission line with total loop resistance 5.0 Ω. Compare this with delivering the same 20.0 kW into the line at 250 V. Treat 20.0 kW as power at the sending end and use rms values. The graph supplies transformer voltage against turns ratio. Supplied model: For an ideal transformer use Vs/Vp = Ns/Np and VpIp = VsIs; wire heating is I²R. Supplied data (Secondary/primary turns ratio; Secondary voltage / V): (1, 250), (10, 2500), (20, 5000), (40, 10000). (a) Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case. [3 marks] (b) Evaluate the direct 250 V comparison and explain why a transformer does not work continuously with steady DC. [3 marks]
Ideal transformer voltage ratio
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Secondary voltage / V
Horizontal axis (x): Secondary/primary turns ratio
Read the exact data as a table
| Secondary/primary turns ratio | Secondary voltage / V |
|---|---|
| 1 | 250 |
| 10 | 2500 |
| 20 | 5000 |
| 40 | 10000 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case.
Reveal the worked solution
- Vs/Vp = Ns/Np = 10000/250 = 40. At the sending end Is = P/Vs = 20000/10000 = 2.0 A.
- For an ideal transformer input power equals output power, so Ip = 20000/250 = 80 A.
- Line heating is I²R = 2²(5) = 20 W, leaving 20000−20 = 19980 W at the receiving end; the line voltage drop is IR = 10 V.
- Using sending current P/V = 20000/250 = 80 A would give a 400 V line drop and I²R = 32000 W, exceeding the sending power.
- This contradicts a passive receiving load at the assumed operating point; the calculation demonstrates why voltage drop and the load must be solved together, and why high-voltage transmission is useful.
- A steady DC current produces steady flux after the switching transient, so it induces no sustained secondary emf. Real AC transformers also have core and winding losses and insulation limits.
Answer (a) Turns ratio = 40; secondary current = 2.0 A; primary current = 80 A; line loss = 20 W; received power = 19.98 kW. (b) The naive 250 V case demands 32 kW of line loss for only 20 kW sent, revealing that the assumed delivered operating point is impossible with a passive load.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.