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A galvanometer gives full scale deflection with $0.006 A$ current. By connecting it to a $4990 Ω$ resistance, it can be converted into a voltmeter of range $0 - 30 V$ . If connected to $a \frac{2 n}{249} Ω$ resistance, it becomes an ammeter of range $0 - 1.5 A$ . The value of $n$ is

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answer is C.

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Detailed Solution

$i_{g} (G + 4990) = V$

$\frac{6}{1000} (G + 4990) = 30$

$G + 4990 = \frac{30000}{6} = 5000$

$G = 10 Ω$

$V_{ω} = V_{e d}$

$i_{g} G = (1.5 - i_{g}) S$

$\frac{6}{1000} \times 10 = (1.5 - \frac{6}{1000}) S$

$S = \frac{60}{1494} = \frac{2 n}{249}$

$n = \frac{249 \times 30}{1494}$

$= \frac{2490}{498} = 5$

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