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 2.97x 10^-2 cm² that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (specific resistance of wire = 5 x 10^-7 $Ω$ m)

Moderate

A

8
B

6
C

3
D

1.5

Solution

We know that $(i - i_{g}) S = i_{g} G$
$∴ i = \frac{(G + S)}{S} i_{g}$
Therefore $5 = \frac{50 + R}{R} \times 0 \cdot 05$
or $R = \frac{2 \cdot 5}{4 \cdot 95}$
Further $R = ρ (\frac{l}{A}) or l = \frac{AR}{ρ}$
or $l = \frac{(2 \cdot 97 \times 10^{- 6}) \times (2 \cdot 5)}{(5 \times 10^{- 7}) \times 4 \cdot 95} = 3 m$

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