A Level Physics 21.2 Rectification and Smoothing MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Cambridge AS and A Level Physics 9702 Section 21.2, Rectification and Smoothing, an A Level (A2) topic in the Alternating Currents unit. Practise 30 original MCQs on half-wave and full-wave rectification, the bridge rectifier and capacitor smoothing, each with the answer and a full explanation.
Key facts recap: Rectification converts AC to DC using diodes, which conduct in one direction only. Half-wave rectification (1 diode) gives one output pulse per cycle, so the pulse frequency equals the supply frequency. A full-wave bridge (4 diodes) uses both half cycles, the load current is always in the same direction, and there are two output pulses per cycle, so the ripple frequency is twice the supply frequency. For smoothing, a capacitor is connected in parallel with the load: it charges to near the peak voltage, then discharges through the load with time constant RC; a larger C or a larger R gives a smaller ripple. See the full summary on the A Level Physics Formula Sheet.
Half-Wave & Full-Wave Rectification
Q1. What is meant by rectification?
- (A) Increasing the frequency of an alternating supply
- (B) Converting an alternating current into a direct current
- (C) Reducing the peak voltage of an alternating supply
- (D) Converting a direct current into an alternating current
Show Answer
Answer: B – Rectification is the conversion of AC into DC. It is done with diodes, which allow current to pass in one direction only, so the output always flows the same way through the load.
Q2. A diode is used in rectifier circuits because it
- (A) stores electric charge
- (B) produces a constant output voltage
- (C) allows current to pass in one direction only
- (D) increases the current in both directions
Show Answer
Answer: C – A diode conducts only when it is forward biased. During the half cycles when it is reverse biased it blocks the current, and this one-way action is what turns AC into a one-directional output.
Q3. How many diodes are used in a half-wave rectifier?
- (A) one
- (B) two
- (C) three
- (D) four
Show Answer
Answer: A – A half-wave rectifier uses a single diode in series with the load. It conducts during one half of each cycle only.
Q4. In a half-wave rectifier, the output voltage across the load resistor is
- (A) a steady direct voltage
- (B) a pulse during every half cycle of the supply
- (C) an alternating voltage at half the supply frequency
- (D) a pulse during alternate half cycles only, and zero during the other half cycles
Show Answer
Answer: D – The single diode conducts for only one half of each cycle. During the other halves it is reverse biased, no current flows, and the output is zero, so the output is very uneven.
Q5. A half-wave rectifier is fed from a 50 Hz AC supply. The output pulses are repeated at a frequency of
- (A) 25 Hz
- (B) 50 Hz
- (C) 100 Hz
- (D) 200 Hz
Show Answer
Answer: B – There is one output pulse in each complete cycle of the supply, so the pulse frequency equals the supply frequency, 50 Hz.
Q6. A full-wave bridge rectifier contains
- (A) one diode
- (B) two diodes
- (C) four diodes
- (D) no diodes, only resistors
Show Answer
Answer: C – The bridge uses four diodes. One pair conducts during one half cycle and the other pair conducts during the next half cycle.
Q7. In a full-wave bridge rectifier, the current through the load resistor is always in the same direction because
- (A) the bridge routes both half cycles of the supply through the load in the same direction
- (B) the supply voltage never reverses
- (C) each diode conducts for the whole cycle
- (D) the load resistor stores charge between half cycles
Show Answer
Answer: A – On each half cycle a different pair of diodes conducts, and the arrangement of the bridge steers the current through the load in the same direction both times, so both half cycles are used.
Q8. A full-wave rectifier is fed from a 50 Hz supply. The unsmoothed output voltage rises to its peak value and falls back to zero at a frequency of
- (A) 25 Hz
- (B) 50 Hz
- (C) 75 Hz
- (D) 100 Hz
Show Answer
Answer: D – Both half cycles produce an output pulse, so there are two pulses per cycle: 2 × 50 Hz = 100 Hz.
Q9. Which description matches the unsmoothed output of a full-wave bridge rectifier?
- (A) A steady voltage equal to the peak value of the supply
- (B) A series of pulses, each rising from zero to the peak value and falling back to zero, with two pulses in each cycle of the supply
- (C) A voltage that is zero for exactly half of each cycle
- (D) A sine wave identical to the input
Show Answer
Answer: B – Every half cycle gives one pulse, so the output is a chain of pulses that each rise from zero to the peak and back to zero. Unlike the half-wave output, there are no long gaps, but the voltage still varies a lot.
Q10. The unsmoothed output of a full-wave rectifier still needs smoothing because
- (A) it is still an alternating current
- (B) it has a mean value of zero
- (C) its voltage still varies, falling to zero at the end of every half cycle
- (D) its pulses are at the wrong frequency
Show Answer
Answer: C – The output is one-directional (DC), but it is pulsating DC: it rises to the peak and falls back to zero twice in each cycle. A steady output needs this variation to be smoothed out.
Smoothing with a Capacitor
Q11. In a smoothed power supply, the smoothing capacitor is connected
- (A) in parallel with the load resistor
- (B) in series with the load resistor
- (C) in series with the AC supply only
- (D) across the AC supply, before the rectifier
Show Answer
Answer: A – The capacitor is placed across (in parallel with) the load, so it can charge up from the rectifier output and then discharge through the load itself.
Q12. As the rectifier output voltage rises towards its peak value, the smoothing capacitor
- (A) discharges through the load
- (B) blocks the output completely
- (C) reverses its charge
- (D) charges up to almost the peak voltage
Show Answer
Answer: D – While the rectifier output is rising and is above the capacitor voltage, the diodes conduct and the capacitor charges rapidly to nearly the peak value.
Q13. When the rectifier output voltage falls below the voltage across the smoothing capacitor, the diodes stop conducting and
- (A) the output voltage falls instantly to zero
- (B) the capacitor discharges through the load, supplying the load current
- (C) the capacitor charges up the AC supply
- (D) the load current stops completely
Show Answer
Answer: B – With the diodes off, the charged capacitor acts as the source: it discharges slowly through the load resistor, keeping the output voltage up until the next peak tops it up again.
Q14. With a suitable smoothing capacitor fitted, the output voltage of the supply
- (A) is exactly constant at all times
- (B) still falls to zero twice in each cycle
- (C) stays close to the peak value, varying slightly with a small ripple
- (D) becomes alternating again
Show Answer
Answer: C – The capacitor holds the output near the peak value. It falls a little as the capacitor discharges and rises again at the next peak; this small up-and-down variation is the ripple.
Q15. Increasing the capacitance of the smoothing capacitor, with the same load, makes the output smoother because
- (A) the capacitor discharges more slowly between peaks, so the ripple is smaller
- (B) the capacitor charges to a much higher peak voltage
- (C) the diodes conduct for a longer part of each cycle
- (D) the frequency of the supply increases
Show Answer
Answer: A – A larger capacitance stores more charge and loses a smaller fraction of it while discharging through the load, so the voltage falls less between peaks and the ripple is smaller.
Q16. The smoothing capacitor discharges through the load with time constant RC. Increasing the load resistance R, so that the load takes a smaller current,
- (A) makes the capacitor discharge faster
- (B) has no effect on the discharge
- (C) increases the ripple
- (D) makes the capacitor discharge more slowly, reducing the ripple
Show Answer
Answer: D – A larger R gives a larger time constant RC, so the capacitor discharges more slowly and its voltage drops less before the next peak. The output is smoother.
Q17. A smoothed supply is connected to a load that draws twice as much current as before. The ripple on the output will
- (A) stay exactly the same
- (B) decrease, because the capacitor charges faster
- (C) increase, because the capacitor loses more charge between peaks
- (D) fall to zero
Show Answer
Answer: C – A larger load current (smaller load resistance) drains the capacitor faster, so its voltage falls further before the next peak tops it up. The ripple therefore increases.
Q18. Smoothing is more effective with a full-wave rectifier than with a half-wave rectifier using the same capacitor and load, because
- (A) the full-wave output has a higher peak voltage
- (B) the capacitor is topped up twice as often, so it has to supply the load for only about half as long between peaks
- (C) the bridge removes the ripple completely
- (D) the load current is always smaller
Show Answer
Answer: B – With full-wave rectification the next peak arrives after half a cycle instead of a full cycle. The capacitor discharges for a shorter time, loses less charge, and the ripple is smaller.
Q19. During most of each cycle in a smoothed supply, the load current is supplied mainly by
- (A) the AC supply, through the diodes
- (B) the smoothing capacitor, as it discharges
- (C) the transformer core
- (D) charge stored in the diodes
Show Answer
Answer: B – The diodes conduct only briefly near each peak, when they top the capacitor up. For the rest of the time the capacitor alone supplies the load as it discharges.
Q20. The smoothing capacitor is recharged (topped up) during each cycle
- (A) at a constant rate throughout the cycle
- (B) only briefly, when the rectifier output voltage rises above the capacitor voltage near each peak
- (C) only at the moment the supply is switched on
- (D) during the half cycles when the diodes are reverse biased
Show Answer
Answer: B – A diode can only conduct when the rectifier output is higher than the capacitor voltage. This happens for a short interval around each peak, and in that interval the capacitor quickly recharges to the peak value.
Rectification Choices & Problems
Q21. A half-wave rectifier is connected to a 60 Hz AC supply. The output pulses are produced at a frequency of
- (A) 60 Hz
- (B) 120 Hz
- (C) 30 Hz
- (D) 240 Hz
Show Answer
Answer: A – A half-wave rectifier gives one pulse per supply cycle, so the pulse frequency equals the supply frequency, 60 Hz.
Q22. A full-wave bridge rectifier is connected to a 60 Hz supply and fitted with a smoothing capacitor. The ripple on the output has a frequency of
- (A) 30 Hz
- (B) 60 Hz
- (C) 120 Hz
- (D) 240 Hz
Show Answer
Answer: C – The capacitor is topped up twice per cycle, once on each half cycle, so the output rises and falls twice per cycle: ripple frequency = 2 × 60 = 120 Hz.
Q23. In a bridge rectifier, one of the four diodes fails and becomes open-circuit. The output across the load becomes
- (A) zero at all times
- (B) a steady direct voltage
- (C) an alternating voltage
- (D) like the output of a half-wave rectifier, with pulses on alternate half cycles only
Show Answer
Answer: D – With one diode open, a conducting path through the load exists for only one half of each cycle; the path needed for the other half cycles is broken. Only alternate half cycles give an output, exactly as in half-wave rectification.
Q24. Two otherwise identical smoothed supplies, using the same load, have smoothing capacitors of 100 μF and 1000 μF. The supply with the 1000 μF capacitor has the smaller ripple because
- (A) its capacitor has a ten times larger time constant RC, so it discharges much less between successive peaks
- (B) its capacitor charges to ten times the voltage
- (C) its diodes conduct ten times as often
- (D) its output frequency is ten times higher
Show Answer
Answer: A – Ripple size depends on how far the capacitor voltage falls while it discharges through the load. Ten times the capacitance gives ten times the time constant, so the voltage falls far less before the next peak.
Q25. The smoothing capacitor is removed from a smoothed full-wave supply. The output voltage across the load will now
- (A) remain steady at the peak value
- (B) become zero at all times
- (C) rise to the peak value and fall back to zero twice in each cycle of the supply
- (D) reverse direction every half cycle
Show Answer
Answer: C – Without the capacitor there is nothing to hold the voltage up between peaks, so the output returns to the unsmoothed full-wave form: a pulse rising to the peak and falling to zero in every half cycle.
Q26. For the same AC supply, which arrangement gives a DC output whose mean (average) value is closest to the peak value of the supply?
- (A) A full-wave rectifier fitted with a large smoothing capacitor
- (B) A half-wave rectifier with no smoothing capacitor
- (C) A full-wave rectifier with no smoothing capacitor
- (D) The AC supply connected directly to the load
Show Answer
Answer: A – Smoothing holds the output near the peak value, and full-wave rectification tops the capacitor up twice per cycle, so the voltage dips least and the mean output stays highest, close to the peak.
Q27. A phone charger takes in AC from the mains and contains a rectifier and a smoothing capacitor. The smoothing stage is needed because
- (A) the phone battery can only be charged by alternating current
- (B) the mains frequency is too high for the phone
- (C) the phone circuits need a nearly steady direct voltage, whereas the rectified output alone rises and falls to zero every half cycle
- (D) the capacitor increases the supply voltage
Show Answer
Answer: C – Electronic devices and batteries need DC at a steady voltage. Rectification gives the correct direction but a pulsating voltage, so the capacitor smooths it to an almost constant level.
Q28. A smoothing capacitor of 2000 μF discharges through a load resistor of 100 Ω. The time constant for the discharge is
- (A) 0.20 s
- (B) 0.050 s
- (C) 2.0 s
- (D) 20 s
Show Answer
Answer: A – Time constant = RC = 100 × 2000 × 10−6 = 0.20 s. This is the time scale on which the capacitor discharges through the load.
Q29. A smoothed half-wave supply needs a much larger capacitor than a smoothed full-wave supply (same load, 50 Hz mains) to give the same small ripple, because in the half-wave supply
- (A) the peak output voltage is much lower
- (B) the diodes conduct in both directions
- (C) the capacitor is charged to the peak only once per second
- (D) the capacitor must supply the load for almost a whole cycle, about 20 ms, between peaks, twice as long as in the full-wave supply, so it loses more charge
Show Answer
Answer: D – At 50 Hz one cycle lasts 20 ms. The half-wave capacitor is topped up once per cycle, so it discharges for about 20 ms; the full-wave capacitor is topped up every 10 ms. To lose the same small amount of charge over the longer interval, the half-wave capacitor must be much larger.
Q30. Which change would reduce the ripple on the output of a smoothed rectifier supply?
- (A) Decreasing the capacitance of the smoothing capacitor
- (B) Decreasing the load resistance so that a larger current is drawn
- (C) Using half-wave rectification instead of full-wave rectification
- (D) Using full-wave rectification together with a larger smoothing capacitor
Show Answer
Answer: D – A larger capacitor discharges less between peaks, and full-wave rectification halves the time between top-ups. Both changes make the capacitor voltage fall less, so the ripple is reduced. The other changes all make the ripple bigger.
Keep practising: Find more A Level topics and every other board on the Physics MCQs hub, and keep the key equations beside you with the A Level Physics Formula Sheet.
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