A galvanometer of resistance \( 8\,\Omega \) is shunted by a resistance of \( 2\,\Omega \). If the total current is \( 1\,\mathrm{A} \), the part of current passing through the shunt will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.8\,\mathrm{A} \)
The galvanometer and the shunt are in parallel, so they have the same potential difference across them, and the current divides in the inverse ratio of their resistances.
\( G = 8\,\Omega, \quad S = 2\,\Omega, \quad I = 1\,\mathrm{A} \)
\( \displaystyle I_S = I\,\frac{G}{G+S} \)
\( \displaystyle I_S = 1\,\mathrm{A} \times \frac{8\,\Omega}{8\,\Omega + 2\,\Omega} \)
\( \displaystyle I_S = 0.8\,\mathrm{A} \)
The galvanometer carries the remaining 0.2 A.
Note: A similar question was also asked in CUET (UG) 2023 on 20-06-2023 (Slot 3).