Question

A galvanometer having a coil resistance of $100 \Omega$ gives a full scale deflection when a current of $1mA$ is passed through it. What is the value of the resistance, which can convert the galvanometer into an ammeter giving a full scale deflection for a current of $10A$?
$\left( A \right)$ $0.01\Omega$
$\left( B \right)$ $2\Omega$
$\left( C \right)$ $0.1\Omega$
$\left( D \right)$ $3\Omega$

Answer 1

Here we are given the resistance and the number of current passes through it.
We can find the value resistance by calculating the maximum voltage and the shunt resistance across the coil, since a galvanometer cannot measure heavy current, a very low shunt resistance is connected parallel to it.

Complete step by step answer:
Here, the value of the shunt resistance is so adjusted that most of the current pass through the shunt.
If the resistance of the galvanometer is $R_g$ it gives full-scale deflection when the current $I_g$ is passed through it then,
$\Rightarrow V = I_g R_g$
$\Rightarrow V = 1mA \times 100\Omega$
Let a shunt of resistance ( $R_s$) be connected in parallel to a galvanometer.
If the total current through the circuit is $I,$ then the value of $I$ is calculated by adding the current in the shunt resistance and the galvanometer.
By using the formula for the voltage we can calculate the value of shunt resistance.
$\Rightarrow I = 10A = I_s + I_g$
$\Rightarrow V = I_s R_s = (I - I_g)R_s$
$\Rightarrow (I - I_g)R_s = I_g R_g$
$\Rightarrow (10 - {10^{ - 3}})R_s = 100 \times {10^{ - 3}}$ $(1m = {10^{ - 3}}).$
Thus the approximate value of shunt resistance will be,
$R_s \approx 0.01\Omega$

Hence the correct option is $\left( A \right)$, thus nullifying other options.

Note: A Galvanometer is a device used to detect feeble electric currents in the circuit.
A coil has been pivoted between the concave pole faces of a strong laminated horseshoe magnet.
When an electric current is passed through the coil, it deflects.
Thus only a small amount of current can be measured by the galvanometer. In order to measure the large current, it has to be converted into an ammeter.
A low resistance called Shunt resistance can be connected in parallel to the galvanometer to convert it into an ammeter.