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WAEC Physics 2010 Paper

Question : 38

Total: 50

\Omega

\Omega

Solution:

Concept:

A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.

Explanation:

Full-scale deflection current is

I_g = 20\ \mathrm{mA} = 0.02\ \mathrm{A}

Total resistance in series is

R_{\text{total}} = 50\ \Omega + 1950\ \Omega = 2000\ \Omega

Maximum voltage is calculated using

V = I_g \times R_{\text{total}}

Thus,

V = 0.02 \times 2000 = 40\ \mathrm{V}

So the maximum voltage readable by the voltmeter is

40\ \mathrm{V}

Answer:

D. 40 V

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