Question

A galvanometer has a resistance of 99 $\Omega$. It is shunted with$1\Omega$ resistance. The ratio of the currents flowing through the galvanometer and through the shunt is-

• A. 1:99
• B. 99:1
• C. 1:100
• D. 100:1

Answer 1

The correct option is A 1:99

Given: Resistance of galvanometer = $99\Omega$

As the galvanometer is shunted, both are connected in parallel.

I(through shunt) = V

I(through galvanometer) = $\frac{V}{99}$

Ratio = $\frac{1}{99}$