If only $1 \%$ of total current is passed through a galvanometer of resistance ' $\mathrm{G}$ ' then the…

If only $1 \%$ of total current is passed through a galvanometer of resistance ' $\mathrm{G}$ ' then the resistance of the shunt is
  1. $\frac{\mathrm{G}}{25} \Omega$
  2. $\frac{\mathrm{G}}{49} \Omega$
  3. $\frac{\mathrm{G}}{2} \Omega$
  4. $\frac{\mathrm{G}}{99} \Omega$

Solution

Given: $I_g=\frac{1}{100} I$ $\begin{aligned} & \therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}=100 \\ & \quad \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{B}}}{\mathrm{I}-\mathrm{I}_8}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}-1} . \end{aligned}$ Substituting, (i) intó (ii), $\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$

Asked in: MHT CET 2023 (10 May Shift 1)

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