Question

If a galvanometer is shunted by \((\frac{1}{n-1})^{th}\) of the value of its resistance, then the fraction of the total current passing through the galvanometer is

• A. \(\frac{1}{(n-1)}\)
• B. \(\frac{1}{n}\)
• C. \(\frac{n}{(n-1)}\)
• D. \(\frac{1}{1-n}\)

Hint:

Currents split in the ratio 1 : (n - 1).

Answer 1

The Correct Option is B

Step 1: Parallel rule:
Current in a branch is proportional to the other branch resistance: $\dfrac{I_g}{I_s}=\dfrac SG=\dfrac1{n-1}$.

Step 2: Total:
$I=I_g+I_s=I_g(1+(n-1))=nI_g$.

Step 3: Result:
$I_g/I=1/n$. Option (B).

Final Answer:
The shunt takes n - 1 parts for every 1 part in the galvanometer. \[ \boxed{B} \]