Question

A Galvanometer has a sensitivity of $60\text{ }division/ampere$. When a shunt is used its sensitivity becomes $10\text{ }division/ampere$. If the Galvanometer is of resistance $20\Omega$. The value of shunt used
A. $4\Omega$
B. $5\Omega$
C. $20\Omega$
D. $2\Omega$

Answer 1

Initial sensitivity and resistance of galvanometer, sensitivity of galvanometer when shunt is connected are given in the question. We need to find the value of shunt resistance. Here we can use the equation for shunt resistance which relates all the above terms.
Formula used:
$S=\dfrac{{{i}_{g}}}{\left( i-{{i}_{g}} \right)}\times G$

Complete answer:
Given,
Galvanometer sensitivity or current through the circuit,$\text{i}=\text{ }60\text{ }division/amp$
Galvanometer sensitivity when shunt is used or galvanometer current,
${{\text{i}}_{g}}=\text{ }10~division/amp$
Resistance of galvanometer, $\text{G}=20\Omega$
We have,
$S=\dfrac{{{i}_{g}}}{\left( i-{{i}_{g}} \right)}\times G$ --------- 1
Where,
${{i}_{g}}$is the Galvanometer current
$i$ is the total current passing through the circuit
$S$ is the Shunt resistance
$G$ is the galvanometer resistance
Substituting the given values in equation 1, we get,
$S=\dfrac{10}{60-10}\times 20$
$S=4\Omega$

So, the correct answer is “Option A”.

Additional Information:
A Galvanometer is used in a circuit to detect the presence of small current and voltage and measure their magnitude. It is mainly used in potentiometers and bridges where it shows null deflection or zero current. It works based on the principle that, when a current sustaining coil is kept in a magnetic field, it experiences a torque. In a Galvanometer a shunt is used to measure heavy current.

Note:
A shunt resistance can be connected in parallel or series to the galvanometer. Galvanometer with a shunt connected in parallel acts as an ammeter. Galvanometer acts as a voltmeter, when the shunt is connected in series to the galvanometer. It also serves the purpose of protecting the Galvanometer from damage due to high current.