The deflection in a moving coil galvanometer is reduced to half when it is shunted with a 40 Ω coil. The resistance of the galvanometer is

Question

The deflection in a moving coil galvanometer is reduced to half when it is shunted with a 40 Ω coil. The resistance of the galvanometer is (A) 80 Ω (B) 40 Ω (C) 20 Ω (D) 15 Ω

Tardigrade Admin · Accepted Answer

We know, (IG/I)=(1/2) (S/G+S)=(1/2) (IG/I)=(40/G+40)= (1/2) G=40 Ω

Q. The deflection in a moving coil galvanometer is reduced to half when it is shunted with a $40 Ω$ coil. The resistance of the galvanometer is

$80 Ω$

$40 Ω$

$20 Ω$

$15 Ω$

We know, $\frac{I _{G}}{I} = \frac{1}{2}$
$\frac{S}{G + S} = \frac{1}{2}$
$\frac{I _{G}}{I} = \frac{40}{G + 40} = \frac{1}{2}$
$G = 40 Ω$

Q. The deflection in a moving coil galvanometer is reduced to half when it is shunted with a 40Ω coil. The resistance of the galvanometer is

80Ω

40Ω

20Ω

15Ω