The resistor value required to use a 0-1 mA meter with an internal resistance of 125 Ω for a 0-1 V meter will be :
875 Ω
To make a voltmeter, a resistor is added in series so that full-scale voltage produces exactly full-scale current.
Step 1 — the total resistance needed. At full scale the meter must draw 1 mA when 1 V is applied across the whole combination:
\(R_{total}=\dfrac{V}{I_{m}}=\dfrac{1}{1\times10^{-3}}=1000\ \Omega\)
Step 2 — subtract what the movement already contributes.
\(R_{s}=R_{total}-R_{m}=1000-125=875\ \Omega\)
— option 3.
Why each wrong option is wrong tells the whole story.
| Option | Where it comes from |
|---|---|
| 1000 Ω | The total resistance — forgetting to subtract Rm |
| 1125 Ω | Adding Rm instead of subtracting it |
| 250 Ω | Unrelated to either quantity |
| 875 Ω | ✓ Correct |
Option 1 is the natural slip: the coil's own 125 Ω is part of the 1000 Ω the circuit must present, so the added resistor supplies only the remainder.
Series for voltage, shunt for current — and the reason is the same in both cases. A voltmeter is connected across the circuit under test, so it must draw as little current as possible, which means a high resistance in series. An ammeter is connected in the circuit, so it must drop as little voltage as possible, which means a low resistance in parallel with the movement.
The figure of merit that follows. A voltmeter's quality is quoted as its sensitivity in ohms per volt, which is simply the reciprocal of the full-scale current:
\(S=\dfrac{1}{I_{m}}=\dfrac{1}{1\ \text{mA}}=1000\ \Omega/\text{V}\)
and the total resistance on any range is \(S\) multiplied by that range. A more sensitive movement — 50 µA, giving 20 000 Ω/V — loads the circuit twenty times less, which is why such movements were prized in analogue multimeters. Loading error matters most in high-impedance circuits, where a 1000 Ω/V meter can change the very voltage it is trying to measure.
Hence, the required series resistance is 875 Ω.
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