A galvanometer having a resistance of \( 8\,\Omega \) is shunted by a wire of resistance \( 2\,\Omega \). If the current is 1 A, the part of it passing through the shunt will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.8 A
A shunt is a small resistance connected in parallel with the galvanometer. The two have the same potential difference, so the current divides in the inverse ratio of the resistances.
\( G = 8\,\Omega \)
\( S = 2\,\Omega \)
\( I = 1\,\mathrm{A} \)
\( \displaystyle I_S = I\,\frac{G}{G + S} = 1 \times \frac{8}{8 + 2} = 0.8\,\mathrm{A} \)
The galvanometer takes the remaining \( 1 - 0.8 = 0.2\,\mathrm{A} \).
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).