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An ammeter has a current range of $0$ to $5\,\text{A}$ and an internal resistance of $0.2\,\Omega$. To change the range to $0$ to $25\,\text{A}$, we need to add a resistance of

1. 0.8 Ω in series with the meter.
2. 1.0 Ω in series with the meter.
3. 0.04 Ω in parallel with the meter.
4. 0.05 Ω in parallel with the meter.

Correct answer: 0.05 Ω in parallel with the meter.

Solution

At full-scale, the meter carries 5 A through 0.2 Ω, so the voltage across it is $V=5\times0.2=1\,\text{V}$. For a 25 A range, the shunt carries $25-5=20\,\text{A}$ at the same 1 V, so $R_s=1/20=0.05\,\Omega$.

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