Correct answer: B — converts energy from the mains into DC for operating electronic equipment

A mains-operated DC power supply takes the 230 V AC supplied by the electricity network and converts it into a steady DC voltage suitable for powering electronic equipment. Typically it does this through four stages: a transformer (to step the voltage up or down), a rectifier (diodes to convert AC to pulsating DC), a filter (capacitors and/or inductors to smooth the pulses), and a regulator (to stabilise the output voltage).

• A is wrong because mains electricity is already AC — a power supply converts AC to DC, not DC to AC. A device that converts DC to AC is called an inverter.
• C is wrong because a diode-capacitor combination is used for rectification and filtering within a power supply, not for measuring mains power. Power is measured with a wattmeter.
• D is wrong because there is no standard "diode-choked device for measuring inductance power" — this description is meaningless in normal electronics practice.

Therefore, a mains-operated DC power supply converts mains AC energy into the DC voltages needed to operate electronic equipment.

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Correct answer: D — a full-wave diode bridge

Rectification is the process of converting alternating current (AC) into pulsating direct current (DC). A full-wave diode bridge consists of four diodes arranged so that both the positive and negative half-cycles of the AC input are steered in the same direction through the load, producing a full-wave rectified output. This is the fundamental rectifying stage in a DC power supply before filtering and regulation.

• A — an electrolytic capacitor is wrong because a capacitor is used for filtering, smoothing the pulsating DC output after rectification, not for converting AC to DC.
• B — a fuse is wrong because a fuse is a protection device that interrupts the circuit when excess current flows; it plays no role in rectification.
• C — a crowbar is wrong because a crowbar circuit is an overvoltage protection device that deliberately short-circuits the supply if the output voltage exceeds a safe threshold; it does not rectify.

Therefore, the component in a DC power supply that performs rectification is the full-wave diode bridge.

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Correct answer: A — an electrolytic capacitor

After rectification, the DC output of a power supply is not smooth — it contains a ripple voltage that rises and falls at the rectifier's output frequency. An electrolytic capacitor connected across the supply output acts as a reservoir: it charges up during each voltage peak and discharges slowly between peaks, filling in the valleys and producing a much smoother DC voltage. Electrolytic capacitors are chosen for this role because they offer high capacitance in a compact size, making them ideal for storing the charge needed to maintain a steady output.

• A fuse is a protection device that interrupts the circuit if current exceeds a safe level; it plays no part in smoothing.
• A crowbar is an overvoltage protection circuit that short-circuits (or "crowbars") the supply if the output voltage rises too high, protecting downstream equipment; it does not smooth ripple.
• A full-wave diode bridge is the rectifier itself — it converts AC to pulsating DC, which is the signal that needs smoothing, not the component that provides it.

Therefore, an electrolytic capacitor is the component in a DC power supply that performs the smoothing operation by filtering out ripple voltage.

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Correct answer: A — a 12 volt car battery

A solid-state VHF transceiver rated at 10 W RF output typically draws around 2–4 A from a 12 V supply (accounting for transmitter efficiency and receiver current). A 12 V car battery can comfortably supply tens of amperes continuously, making it well suited to power such a transceiver on both transmit and receive.

\[ I = \frac{P}{V} = \frac{10}{12} \approx 0.83\ \mathrm{A\ (RF\ output\ only)} \]

In practice, DC input current is higher than this due to efficiency losses — typically 2–4 A on transmit — so the supply must handle that demand without voltage collapse.

• B. 6 penlite cells in series — Six AA (penlite) cells in series give around 9 V and have a very limited current capacity (typically a few hundred mA). This is insufficient voltage and current for a 10 W transceiver.
• C. a 12 volt, 500 mA plug-pack — A 500 mA (0.5 A) plug-pack can only deliver 6 W maximum, and its output voltage would sag badly under the 2–4 A transmit load. Completely inadequate for a 10 W transceiver.
• D. a 6 volt 10 Amp-hour Gel cell — At only 6 V, this supply is below the minimum operating voltage required by a 12 V transceiver. Most such radios will not function correctly or at all below approximately 10–11 V.

Therefore, a 12 V car battery is the only option that provides both the correct voltage and sufficient current capacity to reliably power a 10 W VHF transceiver.

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Correct answer: 100 Hz

New Zealand mains frequency is 50 Hz. In a full-wave rectifier, both the positive and negative half-cycles of the AC waveform are used to produce output pulses.

This means the rectified waveform has two ripple peaks per AC cycle, so the ripple frequency is doubled:

\[ f_{\text{ripple}} = 2 \times f_{\text{mains}} = 2 \times 50\ \mathrm{Hz} = 100\ \mathrm{Hz} \]

• 25 Hz is unrelated to mains rectification.
• 50 Hz would be the ripple frequency for a half-wave rectifier, not a full-wave rectifier.
• 70 Hz does not correspond to any standard relationship with a 50 Hz supply.

Therefore, a full-wave DC power supply operating from New Zealand mains has a ripple frequency of 100 Hz.

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Correct answer: D — 10,000 µF

A smoothing (filter) capacitor in a DC power supply works by charging up during each rectified half-cycle and discharging into the load between peaks, thereby reducing the residual AC ripple voltage. For effective smoothing at mains frequencies (50 Hz in New Zealand), the capacitor must store enough charge to supply the load current during the gaps between rectifier pulses without its voltage dropping significantly.

The required capacitance is determined by the relationship:

\[ C = \frac{I \cdot \Delta t}{\Delta V} \]

For a 1 A load, a 50 Hz full-wave rectifier (pulse interval ≈ 10 ms), and an acceptable ripple of about 1 V:

\[ C = \frac{1 \times 0.010}{1} = 0.01\ \mathrm{F} = 10{,}000\ \mathrm{\mu F} \]

This confirms that a capacitor in the thousands-of-microfarads range is needed — which is why large electrolytic capacitors are used in mains power supplies.

• 100 pF — Far too small; suitable for RF bypass or tuning circuits, not power-supply smoothing.
• 10 nF — Still orders of magnitude too small; used for high-frequency decoupling, not low-frequency ripple reduction.
• 100 nF — Commonly used as a high-frequency bypass capacitor on circuit boards, but provides negligible smoothing at 50 Hz power-supply ripple frequencies.

Therefore, 10,000 µF is the only value large enough to effectively smooth the output of a 1 A, 12 V DC power supply operating from a 50 Hz mains rectifier.

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There should always be a fuse in the phase or active AC mains lead for protection if a fault develops in the equipment. The fuse should be of the correct rating for the task. Keep some spare fuses handy!

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Correct answer: 50 Hz

In New Zealand, the mains supply frequency is:

\[ 50\ \mathrm{Hz} \]

A half-wave rectifier uses only one half of each AC cycle to produce the DC output.

Therefore, the ripple frequency is the same as the mains frequency:

\[ f_{\text{ripple}} = 50\ \mathrm{Hz} \]

• 25 Hz would require additional frequency division.
• 70 Hz is unrelated to the mains supply.
• 100 Hz would be the ripple frequency from a full-wave rectifier.

Therefore, the ripple frequency is 50 Hz.

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Correct answer: all power supplies have some internal resistance

A real DC power supply is not ideal and has some internal resistance.

When current is drawn:

• a voltage drop occurs inside the supply
• this reduces the output voltage available at the terminals

This follows Ohm’s Law:

\[ V_{\text{drop}} = I R_{\text{internal}} \]

• Mains voltage and frequency do not change due to load in this way.
• Power is not reflected back into the mains.

Therefore, the output voltage decreases because all power supplies have some internal resistance.

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Correct answer: D — they can be obtained in larger values than other types

Electrolytic capacitors use a very thin oxide layer as the dielectric, formed electrochemically on aluminium or tantalum foil. This extremely thin dielectric allows very high capacitance values — typically from a few microfarads up to tens of thousands of microfarads — in a physically compact package. Power supply filter and reservoir circuits require large capacitance values to smooth the rectified AC into steady DC, making electrolytics the practical choice for this role.

• A — Capacitors are not "tuned" to a frequency; they respond to all frequencies according to their capacitive reactance. Electrolytics are actually poor performers at high frequencies due to internal inductance and resistance.
• B — Electrolytic capacitors have relatively high losses (expressed as ESR — equivalent series resistance) compared to film or ceramic types. Low-loss types such as polypropylene film capacitors are preferred where low loss matters.
• C — Electrolytics have no special RF noise-shielding property. Their poor high-frequency characteristics actually make them less useful for RF bypassing, which is why small ceramic capacitors are often placed in parallel with them.

Therefore, electrolytic capacitors are chosen for power supply filtering because their construction allows them to achieve the large capacitance values needed to smooth rectified DC effectively.

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