Question

A galvanometer having a resistance of 8 $\Omega$ is shunted by a wire of resistance 2 $\Omega$. If the total current is 1 A, the part of it passing through the shunt will be

• A. 0.25 A
• B. 0.8 A
• C. 0.2 A
• D. 0.5 A

Answer 1

Answer: (B)

0.8 A

Answer 2

Here, $G = 8 \Omega, S = 2 \Omega$ and $I = 1\, A$.

Current through shunt, $I_s = I \frac{G}{(G + S)}$
$= 1 \times \frac{8}{(8 + 2)} = 0.8 A$