Q175 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A galvanometer of resistance '$G$', is shunted by a resistance '$S$' ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
- A$\dfrac{SG}{S+G}$
- B$\dfrac{G}{S+G}$
- Correct answerC$\dfrac{G^2}{S+G}$
- D$\dfrac{S^2}{S+G}$
Solution
Let the original resistance of the galvanometer be $G$.
When the galvanometer is shunted by a resistance $S$, the equivalent resistance of the parallel combination is $\dfrac{GS}{G+S}$.
Let $R$ be the resistance connected in series with this combination.
To keep the main current in the circuit unchanged, the total resistance of the modified galvanometer must be equal to its original resistance $G$.
$R + \dfrac{GS}{G+S} = G$
$R = G - \dfrac{GS}{G+S}$
$R = \dfrac{G(G+S) - GS}{G+S}$
$R = \dfrac{G^2 + GS - GS}{G+S}$
$R = \dfrac{G^2}{S+G}$
Answer: $\dfrac{G^2}{S+G}$
When the galvanometer is shunted by a resistance $S$, the equivalent resistance of the parallel combination is $\dfrac{GS}{G+S}$.
Let $R$ be the resistance connected in series with this combination.
To keep the main current in the circuit unchanged, the total resistance of the modified galvanometer must be equal to its original resistance $G$.
$R + \dfrac{GS}{G+S} = G$
$R = G - \dfrac{GS}{G+S}$
$R = \dfrac{G(G+S) - GS}{G+S}$
$R = \dfrac{G^2 + GS - GS}{G+S}$
$R = \dfrac{G^2}{S+G}$
Answer: $\dfrac{G^2}{S+G}$
Hence, the correct answer is option: 'C'.