MHT CET · Physics · Current Electricity
Only \(4 \%\) of the total current in the circuit passed through a galvanometer. If the resistance of the galvanometer is \(\mathrm{G}\), then the shunt resistance connected to the galvanometer is
- A \(\frac{\mathrm{G}}{25}\)
- B \(\frac{\mathrm{G}}{24}\)
- C \(24 \mathrm{G}\)
- D \(25 \mathrm{G}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{G}}{24}\)
Step-by-step Solution
Detailed explanation
Relationship between the shunt resistance and galvanometer resistance is
\(\mathrm{S}=\frac{\mathrm{I}_0 \times \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}\) ....(i)
\(\mathrm{I}_{\mathrm{g}}=\frac{4}{100} \mathrm{I}=0.04 \mathrm{I}\)
\(\therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=25\) ....(ii)
Putting (ii) into (i)
\(\mathrm{S}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1}=\frac{\mathrm{G}}{25-1}=\frac{\mathrm{G}}{24}\)
\(\mathrm{S}=\frac{\mathrm{I}_0 \times \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}\) ....(i)
\(\mathrm{I}_{\mathrm{g}}=\frac{4}{100} \mathrm{I}=0.04 \mathrm{I}\)
\(\therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=25\) ....(ii)
Putting (ii) into (i)
\(\mathrm{S}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1}=\frac{\mathrm{G}}{25-1}=\frac{\mathrm{G}}{24}\)
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