Question

A galvanometer has a current range of \(10\) mA and a voltage range of \(0.75\) V. To convert this galvanometer into an ammeter of range \(10\) A, what is the shunt resistance?

• A. \(\frac{100}{999}\Omega\)
• B. \(\frac{50}{999}\Omega\)
• C. \(\frac{200}{999}\Omega\)
• D. \(\frac{75}{999}\Omega\)

Hint:

Galvanometer resistance is $G=V/I_g$, and $S=\frac{I_gG}{I-I_g}$.

Answer 1

The Correct Option is D

Step 1: Galvanometer resistance
\(G=\frac{0.75}{10\times10^{-3}}=75\,\Omega\).

Step 2: Shunt
\(S=\frac{I_gG}{I-I_g}=\frac{0.01\times75}{10-0.01}=\frac{0.75}{9.99}=\frac{75}{999}\,\Omega\). Option (D).

Final Answer:
The shunt is \(\frac{75}{999}\,\Omega\), option (D). \[ \boxed{\text{(D)}} \]