A galvanometer has a resistance of 406 Ω. A shunt is connected across it so that its current range becomes 30 times larger. The resistance of the shunt is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 14\ \Omega \)
A shunt is a small resistance joined in parallel with the galvanometer. Most of the current then passes through the shunt, and only the full-scale current \( I_g \) passes through the coil. The coil and the shunt have the same potential difference across them.
The new range is \( I=30I_g \), so the current through the shunt is \( I-I_g=29I_g \).
\( \begin{aligned} I_gG &= (I-I_g)S \\ S &= \frac{I_gG}{I-I_g} \\ &= \frac{I_g\times406}{29I_g} \\ &= 14\ \Omega \end{aligned} \)
Therefore, a shunt of \( 14\ \Omega \) in parallel lets the galvanometer measure currents up to 30 times its full-scale current.
NCERT (2026-27):
The Moving Coil Galvanometer – “… To overcome these difficulties, one attaches a small resistance rs, called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt.”