A-Level · A2 · extended response
An rms reading and a smoothing capacitor
A Level extension · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find rms voltage, rms current and average resistor power. Explain why the zero mean voltage does not imply zero power.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
For a sinusoid, Vrms = Vpeak/√2 = 12/√2≈8.49 V. Irms = Vrms/R≈1.41 A.
- 2
Mean heating power is Vrms²/R = (12²/2)/6 = 12 W. Using peak voltage directly would double this average.
- 3
Voltage and current reverse together, so instantaneous resistor power v²/R stays nonnegative even though the full-cycle mean voltage is zero.
(b)
Find minimum smoothing capacitance for a 0.50 V approximate ripple, and explain the effect of half-wave rectification at the same load.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
A full-wave rectifier provides two charging peaks per 50 Hz cycle, so the time between peaks is approximately 0.010 s.
- 2
C≈IΔt/ΔV = 0.020(0.010)/0.50 = 4.0×10⁻⁴ F = 400 μF.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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