Question

The shunt resistance required to allow $4\%$ of the main current through the galvanometer of resistance $48 \, \Omega$ is

• A. 1 $\Omega$
• B. 2 $\Omega$
• C. 3 $\Omega$
• D. 4 $\Omega$

Answer 1

Answer: (B)

2 $\Omega$

Answer 2

Required shunt $S =\frac{i_{g}}{\left(i-i_{g}\right)} G$ $=\frac{\frac{4}{100} \times i}{\left(i-\frac{4}{100} \times i\right)}=\frac{\frac{i}{25}}{\frac{24}{25} i} \times 48$ $S=2 \Omega$