Correct answer: A — value of any resistance placed between its terminals

An ohmmeter is a test instrument designed specifically to measure electrical resistance. It works by applying a small known voltage or current to the component under test and measuring the resulting current or voltage to calculate the resistance using Ohm's Law. The reading is displayed directly in ohms (Ω), kilohms (kΩ), or megohms (MΩ).

• B is wrong because impedance includes both resistive and reactive (capacitive/inductive) components and varies with frequency. An ohmmeter applies DC, so it cannot measure AC impedance — an impedance bridge or LCR meter is needed for that.
• C is wrong because power factor is a ratio relating true power to apparent power in an AC circuit. Measuring it requires an AC source and appropriate instruments, not a simple ohmmeter.
• D is wrong because an ohmmeter does not measure voltage — that is the function of a voltmeter. An ohmmeter must never be connected to a live circuit, as external voltage will damage it or give false readings.

Therefore, an ohmmeter measures only resistance — the value of any resistance placed between its terminals.

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Correct answer: relative reflected voltage

An SWR/VSWR meter measures forward and reflected signals in a transmission line.

When switched to:

• forward → indicates relative forward voltage (or power)
• reverse → indicates relative reflected voltage

Reflected voltage corresponds to the portion of the signal that is not absorbed by the antenna and is returned toward the transmitter.

• It does not directly display power in watts.
• Forward voltage is measured in the forward position.
• It does not directly measure reflected power in dB.

Therefore, the reverse position indicates relative reflected voltage.

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Correct answer: C — ammeter

An ammeter measures the flow of electric current through a circuit. To measure the supply current to an amplifier, the ammeter is connected in series with the supply line so that all the current flows through it. This gives a direct reading of how many amperes the amplifier is drawing from the power supply.

• A. wattmeter — measures electrical power (watts), not current directly. Power requires both voltage and current to be measured together.
• B. voltmeter — measures the potential difference (voltage) across two points; it is connected in parallel, not in series, and gives no direct reading of current.
• D. ohmmeter — measures resistance (ohms) and must only be used on de-energised circuits; it cannot measure operating supply current.

Therefore, an ammeter connected in series with the supply line is the correct instrument for measuring the current drawn by an amplifier.

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Correct answer: a DC ammeter

A small hand-held transistorised receiver operates from a DC power supply (such as a battery).

To measure the current drawn from this supply, a DC ammeter is connected in series with the power supply line.

• A power meter measures RF output power.
• An RF ammeter measures radio-frequency current.
• An electrostatic voltmeter measures voltage, not current.

Therefore, the correct instrument is a DC ammeter.

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Correct answer: a low value resistance

When measuring current, an ammeter is connected in series with the circuit.

A good ammeter has:

• very low internal resistance
• minimal effect on circuit current

Therefore, in the circuit it behaves like a small (low value) resistance.

• It is not an insulator (which would block current).
• It is not a perfect conductor (zero resistance is ideal but not real).
• While it does draw a tiny amount of power, that is not its intended function.

Therefore, the meter acts as a low value resistance.

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Correct answer: in series with one of the receiver power leads

To measure current, the meter must be placed in series with the load so that all of the supply current flows through the meter.

In a receiver power circuit, this is done by opening one of the power leads (positive or negative) and inserting the ammeter in series at that point. The same current flows in both leads, so measuring either lead gives the correct value.

• in parallel with both receiver power supply leads would short the supply and could damage the meter or power supply.
• in parallel with one of the receiver power leads would not measure load current and may create a short circuit.
• in series with both receiver power leads is unnecessary and impractical, since the same current already flows through each lead.

Therefore, the current meter should be placed in series with one of the receiver power leads.

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Correct answer: A — the resulting high current will probably destroy the ammeter

An ammeter is designed to measure current flowing through a circuit and has a very low internal resistance — often just a fraction of an ohm. Connecting it directly across a 12 V battery places it in a near short-circuit condition. With almost no resistance limiting the current, Ohm's Law predicts an extremely large current will flow, almost certainly burning out the meter movement or blowing the internal fuse.

\[ I = \frac{V}{R} \]

For example, if the ammeter has an internal resistance of 0.01 Ω:

\[ I = \frac{12}{0.01} = 1200\ \mathrm{A} \]

Even a much higher internal resistance of 1 Ω gives 12 A — well beyond the rating of a typical ammeter.

• B is wrong — current will absolutely flow; the battery provides an EMF and the ammeter provides a conductive path.
• C is wrong — 12 V is more than sufficient to drive a damaging current through the low-resistance meter; it is too easy for current to flow, not too hard.
• D is wrong — 12 V is not too high to measure in a general sense; the problem is the near-zero resistance of the ammeter, not an excessively high voltage.

Therefore, connecting an ammeter directly across a battery terminals bypasses any current-limiting resistance and produces a destructive short-circuit current through the meter.

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Correct answer: a very low internal resistance

An ammeter is connected in series with the circuit to measure current flow.

To avoid affecting the current it is measuring, the ammeter must introduce as little additional resistance as possible.

A very low internal resistance ensures:

• minimal voltage drop across the meter

• negligible change to the circuit current

• A high or infinite resistance would significantly reduce current flow.

• Matching the resistance of other components is unnecessary.

Therefore, a good ammeter should have a very low internal resistance.

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Correct answer: A — a very high internal resistance

A voltmeter is connected in parallel with the component whose voltage is being measured. If the meter draws significant current, it loads the circuit and lowers the voltage at that point, giving a reading lower than the true value. A very high internal resistance minimises the current drawn by the meter, so the circuit is barely disturbed and the reading is accurate.

• B. a resistance equal to that of all other components — This would cause the meter to act as a parallel load, halving (or otherwise significantly changing) the voltage across the component under test.
• C. a very low internal resistance — This describes a good ammeter, not a voltmeter. A low-resistance voltmeter would short-circuit the component being measured, drawing large current and giving a wildly incorrect reading.
• D. an inductive reactance — Reactance is frequency-dependent and would make the meter useless for DC measurements; it has no place in a practical voltmeter design.

Therefore, a good voltmeter must have a very high internal resistance so that it measures voltage without significantly loading or disturbing the circuit under test.

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Correct answer: C — 10 volt

An rms-reading voltmeter displays the Root Mean Square (RMS) value of an AC signal. For a pure sinewave, the RMS value is the peak voltage divided by √2 (approximately 1.414). This is the value that represents the equivalent DC heating effect of the waveform.

\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \]

Given a peak voltage of 14 V:

\[ V_{\text{rms}} = \frac{14}{1.414} \approx 9.9\ \mathrm{V} \approx 10\ \mathrm{V} \]

• A. 14 volt — This is the peak voltage, not the RMS value. A peak-reading instrument would show this.
• B. 28 volt — This would be the peak-to-peak voltage (2 × 14 V), which is not what an rms-reading voltmeter displays.
• D. 50 volt — This is the signal frequency in Hz, not a voltage value; it has no relevance to the meter reading.

Therefore, an rms-reading voltmeter measures the effective (heating) value of the sinewave, which is the peak voltage divided by √2, giving approximately 10 V.

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