Question

A galvanometer of resistance 20 $\Omega$ is to be shunted so that only 1% of the current passes through it. Shunt connected is

• A. 99/20$\Omega$
• B. 9/20$\Omega$
• C. 20/99$\Omega$
• D. 2/99$\Omega$

Answer 1

Answer: (C)

20/99$\Omega$

Let stunt connected be R $\Omega$ in parallel

$Req=\frac{20\times R}{20+R}$

$\frac{i}{100}\left(20\right)$ = $\left(\frac{20R}{20+R}\right)i$

As only 1% current is to be passed

= 20+R=100R

99R=20

$R=\frac{20}{99}\Omega$