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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

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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