First slide

Moving coil Galvanometer

Question

A galvanometer, having a resistance of $50 Ω$ , gives a full scale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section $2 .97 \times 10^{- 2} {cm}^{2}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is
(Specific resistance of the wire $= 5 \times 10^{- 7} Ω m)$

Easy

A

9
B

6
C

3
D

1.5

Solution

$\begin{array}{l} \frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{5}{0 .05} = 1 = \frac{50}{S} \\ \Rightarrow S = \frac{50}{99} = \frac{ρ \times l}{A} \Rightarrow l = \frac{50}{99} \times \frac{2 .97 \times 10^{- 2} \times 10^{- 4}}{5 \times 10^{- 7}} = 3 m \end{array}$

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