MHT CET · Physics · Current Electricity
A galvanometer of resistance \(\mathrm{G}\) is shunted with a resistance of \(10 \%\) of \(\mathrm{G}\). The part of the total current that flows through the galvanometer is
- A \(\frac{1}{11} \mathrm{I}\)
- B \(\frac{2}{11} \mathrm{I}\)
- C \(\frac{1}{10} \mathrm{I}\)
- D \(\frac{1}{5} \mathrm{I}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{11} \mathrm{I}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}} & =\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}=\frac{0.1 \mathrm{G}}{0.1 \mathrm{G}+\mathrm{G}}=\frac{1}{11} \\ \therefore \quad \mathrm{I}_{\mathrm{g}} & =\frac{1}{11} \mathrm{I}\end{aligned}\)
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