Question

If resistance of galvanometer is $10\Omega$ and maximum current $i_g$ is 10mA then the shunt resistance required so that the main current 'I' can be up to 1A is $\left(in\Omega \right)$

• A. $\frac{99}{10}$
• B. $\frac{10}{99}$
• C. $990$
• D. $\frac{99}{1000}$

Answer 1

The correct option is B $\frac{10}{99}$

Let, the shunt resistance required to measure the main current 'I' upto 1A in galvanometer is Rs.

Equating potential difference in both cases,

Potential difference across galvanometer = Potential difference across S.

$i_{g}G=(i-i_g)R_s$

$10\times {10}^{-3}\times 10=(1-10\times {10}^{-3})R_s$

$\therefore Rs=\frac{{10}^{-1}}{1-{10}^{-2}}$

$\therefore Rs=\frac{10}{99}\Omega$