MHT CET · Physics · Current Electricity
In an ammeter, \(4 \%\) of the main current is. passing through the galvanometer, If shunt resistance is \(5 \Omega\), then resistance of galvanometer will be
- A \(60 \Omega\)
- B \(120 \Omega\)
- C \(240 \Omega\)
- D \(480 \Omega\)
Answer & Solution
Correct Answer
(B) \(120 \Omega\)
Step-by-step Solution
Detailed explanation
Shunt \(S=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\)
On substituting the given values, we get,
\(\begin{aligned}
\therefore \quad & 5=\frac{\left(\frac{4}{100} \mathrm{I} \times \mathrm{G}\right)}{\mathrm{I}-\frac{4}{100} \mathrm{I}}=\frac{4 \mathrm{G}}{96} \\
& \Rightarrow \mathrm{G}=\frac{96 \times 5}{4}=120 \Omega
\end{aligned}\)
On substituting the given values, we get,
\(\begin{aligned}
\therefore \quad & 5=\frac{\left(\frac{4}{100} \mathrm{I} \times \mathrm{G}\right)}{\mathrm{I}-\frac{4}{100} \mathrm{I}}=\frac{4 \mathrm{G}}{96} \\
& \Rightarrow \mathrm{G}=\frac{96 \times 5}{4}=120 \Omega
\end{aligned}\)
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