Question

A galvanometer gives full scale deflection with $0.2\, A$ in a coil shown in the figure. The resistance of its coil is $ 10\,\Omega $ . How much shunt resistance is required to convert it into an ammeter to read current upto $1.8\, A$ ?

• A. $ 1.25\,\Omega $
• B. $ 2.5\,\Omega $
• C. $ 5.0\,\Omega $
• D. $ 10.0\,\Omega $

Answer 1

Answer: (A)

Potential difference across galvanometer and shunt is same.

For $G$ being resistance of galvanometer and $I_{g}$ the current across it, the current across $S$ is $\left(I-I_{g}\right)$.
Since, $G$ and $S$ are in parallel, potential across them is same
$I_{g} \times G=\left(I-I_{g}\right) \times S $
$\Rightarrow S=\frac{I_{g} G}{\left(I-I_{g}\right)} $
Given, $ I_{g}=0.2 \,A, $
$G=10 \Omega, $
$I=1.8\, A $
$\Rightarrow S=\frac{0.2}{1.6} \times 10$
$=1.25 \,\Omega $