Question

The sensitivity of a galvanometer is $60\,div/A$. When a shunt is used, its sensitivity becomes $10\,div/A$. If the galvanometer is of resistance $20\Omega$, the value of shunt used is:
A. 4 $\Omega$
B. 5 $\Omega$
C. 20 $\Omega$
D. 2 $\Omega$

Answer 1

A galvanometer is an electric current measurement instrument that is electromechanical. Early galvanometers were uncalibrated, but later models, such as ammeters, were calibrated and could more accurately calculate current flow. Galvanometer is a device that uses the deflection of a moving coil to measure a small electrical current or a property of the current. The deflection is a mechanical rotation arising from the current's powers.

Complete step by step answer:
The current in microamperes needed to consume one millimetre deflection on a scale positioned 1 metre away from the mirror is known as the sensitivity of a galvanometer. If the current (I) q is small, the sensitivity would be high. As a result, the galvanometer's sensitivity can be improved by increasing the number of turns (N), using stronger magnets, or increasing the coil's region (A)

The sensitivity of a galvanometer is 60 div/A. When a shunt is used, its sensitivity becomes 10 div/A. Given the galvanometer is of resistance 20 $\Omega$. The galvanometer's sensitivity can be expressed as
$\dfrac{{{{\mathbf{i}}_{\mathbf{g}}}}}{{\mathbf{i}}} = \dfrac{{\mathbf{S}}}{{{\mathbf{S}} + {\mathbf{G}}}}$
$S$ = shunt and $G$ = Galvanometer.
According to the question,

\Rightarrow \dfrac{{{{\mathbf{i}}_{\text{g}}}}}{{\text{i}}}= \dfrac{1}{6}$$ $$\Rightarrow G =20 \Omega $$ Upon substituting $$ \Rightarrow \dfrac{1}{6} = \dfrac{{\mathbf{S}}}{{{\mathbf{S}} + {\mathbf{20}}}}$$ $$ \therefore {\mathbf{S}} = 4\Omega $$ **Hence option A is correct.** **Note:** A shunt is a mechanism in electronics that provides a low-resistance path for electric current to travel through another point in the circuit. The word comes from the verb 'to shunt,' which means to turn around or take a different direction.