MHT CET · Physics · Current Electricity
If only \(1 \%\) of total current is passed through a galvanometer of resistance ' \(\mathrm{G}\) ' then the resistance of the shunt is
- A \(\frac{\mathrm{G}}{25} \Omega\)
- B \(\frac{\mathrm{G}}{49} \Omega\)
- C \(\frac{\mathrm{G}}{2} \Omega\)
- D \(\frac{\mathrm{G}}{99} \Omega\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{G}}{99} \Omega\)
Step-by-step Solution
Detailed explanation
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),
\(\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}\)
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}=100 \\
& \quad \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{B}}}{\mathrm{I}-\mathrm{I}_8}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}-1} .
\end{aligned}\)
Substituting, (i) intó (ii),
\(\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}\)
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