Question

A galvanometer of resistance 240 $\Omega$ allows only 4% of the main current after connecting a shunt resistance. The value of the shunt resistance is

• A. 10 $\Omega$
• B. 20 $\Omega$
• C. 8 $\Omega$
• D. 5 $\Omega$

Hint:

Shunt resistance $S = \frac{I_G G}{I - I_G}$.

Answer 1

The Correct Option is A

Step 1: Given Values
$G = 240~\Omega$. The galvanometer current $I_G = \frac{4}{100}I$.
Step 2: Shunt Formula
The voltage across the galvanometer and shunt are equal: $(I - I_G)S = I_G G$.
Step 3: Calculation
$(I - 0.04I)S = 0.04I \times 240 \Rightarrow 0.96IS = 9.6I \Rightarrow S = \frac{9.6}{0.96} = 10~\Omega$.
Final Answer: (a)