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Q.

A galvanometer of resistance $100 Ω$ gives a full scale deflection for current of $10^{- 5} A$ . To convert it into a ammeter capable of measuring upto 1 A, we should connect a resistance of

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a

$100 Ω$ in series

b

$1 Ω$ in parallel

c

$10^{- 3} Ω$ in parallel

d

$10^{5} Ω$ in series

answer is B.

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Detailed Solution

Here, $R_{g} = 100 Ω; I_{g} = 10^{- 5} A; I = 1 A; S = ?$

$S = \frac{I_{g} R_{g}}{I - I_{g}} = \frac{10^{- 5} \times 100}{1 - 10^{- 5}} = 10^{- 3} Ω$ in parallel

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