In order to pass 10% of main current through a moving coil galvanometer of resistance 99 Ω, the resistance of the required shunt is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 11 Ω
A shunt is a small resistance connected in parallel with the galvanometer, so that most of the current bypasses the coil. Both have the same potential difference across them.
\( I_g = 0.1I \)
\( I_s = I - I_g = 0.9I \)
\( I_g G = I_s S \)
\( \displaystyle S = \frac{I_g G}{I_s} \)
\( \displaystyle S = \frac{0.1I \times 99\,\Omega}{0.9I} \)
\( S = 11\,\Omega \)
Shortcut: With \( n = I/I_g = 10 \), the shunt is \( S = G/(n - 1) = 99/9 = 11\,\Omega \).