Question

A moving coil galvanometer has 50 turns and each turn has an area 2 $\times 10^{-4}$ \[{{m}^{2}}\]. The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is \[{{10}^{-4}}\] Nm rad. When a current flows through the galvanometer, a full-scale deflection occurs if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50 Ω. This galvanometer is to be converted into an ammeter capable of measuring current in the range (0 – 1.0) A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is..........?

Answer 1

Hint: For converting a moving coil galvanometer in to an ammeter we need to connect a low resistance called shunt in parallel with it and if we need to convert moving coil galvanometer into voltmeter, we need to connect a high resistance in series with it.

Complete step by step answer:
Given values are,
N= 50
A= 2 $\times 10^{-4}$ \[{{m}^{2}}\]
B= 0.02 T
K= \[{{10}^{-4}}\]
\[\alpha \]= 0.2 rad
\[{{R}_{g}}\]= 50\[\Omega \]
For equilibrium of the coil, deflecting torque = restoring torque
\[K\alpha =B{{I}_{g}}NA\]
Finding out the value of \[{{I}_{g}}\],
\[{{I}_{g}}=\dfrac{K\alpha }{BNA}=\dfrac{{{10}^{-4}}\times 0.2}{50\times 2\times {{10}^{-4}}\times 0.02}=0.1A\]
\[S=\dfrac{{{I}_{g}}{{R}_{g}}}{I-{{I}_{g}}}=\dfrac{0.1\times 50}{1-0.1}=5.56\Omega \].

Thus, the value of shunt resistance to be connected is 5.56\[\Omega \].

Note:
Deflection of the coil is directly proportional to the current flowing through it. Hence, we use a linear scale in the galvanometer to detect the current flowing in the current. We know the coil moves under the influence of a magnetic field when it passes through it, so using this condition the torque experienced by the coil is given by, \[\tau =BINA\sin \theta \].