A galvanometer of resistance \(100 \Omega\) gives a full scale deflection for a current of \(10^{-5} \mathrm{~A}\).
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A galvanometer of resistance \(100 \Omega\) gives a full scale deflection for a current of \(10^{-5} \mathrm{~A}\). To convert it into an ammeter capable of measuring upto \(1 \mathrm{~A}\), we should connect a resistance of

(a) \(1 \Omega\) in parallel

(b) \(10^{-3} \Omega\) in parallel

(c) \(10^{5} \Omega\) in series

(d) \(100 \Omega\) in series
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The correct option of this question will be (b)

Solution —

$G=100 \Omega, I_{g}=10^{-5} \mathrm{~A}, I=1 \mathrm{~A}$ To convert the galvanometer into an ammeter, we should connect a resistance $S$ in parallel to it.

$\therefore I_{g} \times G=\left(I-I_{g}\right) \times S $

$S=\left(\frac{I_{g}}{I-I_{g}}\right) \times$ $G=\frac{10^{-5}}{1-10^{-5}} \times 100$

or $S=\frac{10^{-3}}{1-0.00001}=10^{-3} Ω$

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