Question

A galvanometer with its coil resistance 25 requires a current of 1 mA for its full deflection. In order to construct an ammeter to read up to a current of 2 A, the approximate value of the shunt resistance should be

Answer 1

Hint: Let G be the resistance of the galvanometer and ${i_g}$ be the current for full scale deflection in the galvanometer, the value of the shunt resistance, S required to convert the galvanometer into an ammeter of 0 to i ampere is,
$S = \dfrac{{G \times {i_g}}}{{i - {i_g}}}$.

Complete step by step answer:
Given the resistance of the galvanometer, G$ = 25\Omega $.
Desired current, i$ = 2A$.
Full scale deflection current, ${i_g} = 1mA$.
To convert a galvanometer into ammeter shunt resistance ‘S’ is added in parallel to the galvanometer.
The value of the shunt resistance is given as, $S = \dfrac{{G \times {i_g}}}{{i - {i_g}}}$.
Substituting the values of G, i and ${i_g}$ we get,
$S = \dfrac{{25 \times 0.001}}{{(2 - 0.001)}} = \dfrac{{0.025}}{{1.999}} = 0.0125 = 1.25 \times {10^{ - 2}}\Omega $

Therefore the value of shunt resistance is $1.25 \times {10^{ - 2}}$.

Additional information:
Galvanometer is a device which is used to measure electric currents. But it is a sensitive device that cannot be used to measure heavy currents. So we connect a resistor in parallel to the galvanometer to measure heavy currents and it is called an Ammeter. The effective resistance when connected in parallel is low.

Note: A galvanometer can detect only small currents. Thus, to measure large currents it is converted into an ammeter. It can be converted into an ammeter by connecting a low resistance called shunt resistance in parallel to the galvanometer. This shunt prevents the galvanometer from damage as it is a sensitive device.