MHT CET · Physics · Current Electricity
When a galvanometer is shunted by a resistance. 's', its current capacity increases ' \(n\) ' times. If the same galvanometer is shunted by another resistance ' \(\mathrm{s}_1\) ', its capacity will increase to ' \(\mathrm{n}_1\) ' times original current. The value of ' \(n_1\) ' is
- A \(\frac{(\mathrm{n}+\mathrm{s})}{\mathrm{s}_1}\)
- B \(\frac{\mathrm{s}_1(\mathrm{n}-\mathrm{s})-\mathrm{s}_1}{\mathrm{~s}_1}\)
- C \(\frac{(\mathrm{n}+1) \mathrm{s}}{\mathrm{s}_1}\)
- D \(\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1}\)
Step-by-step Solution
Detailed explanation
For galvanometer shunt resistance is given by,
\(s=G\left(\frac{1}{1-I / I_g}\right)\)
Current capacity is given by, \(\mathrm{n}=\mathrm{I} / \mathrm{Ig}\)
\(\begin{aligned}
& \mathrm{s}=\mathrm{G}\left(\frac{1}{1-\mathrm{n}}\right) \\
& \therefore \quad \mathrm{G}=\mathrm{s}(\mathrm{n}-1)...(i)
\end{aligned}\)
When same galvanometer is shunted with resistance ' \(\mathrm{s}_1\) '
\(\begin{aligned}
& \mathrm{s}_1=\mathrm{G}\left(\frac{1}{1-\mathrm{n}_1}\right) \\
& \mathrm{n}_1=\frac{\mathrm{G}+\mathrm{s}_1}{\mathrm{~s}_1}=\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1}
\end{aligned}\)
\(s=G\left(\frac{1}{1-I / I_g}\right)\)
Current capacity is given by, \(\mathrm{n}=\mathrm{I} / \mathrm{Ig}\)
\(\begin{aligned}
& \mathrm{s}=\mathrm{G}\left(\frac{1}{1-\mathrm{n}}\right) \\
& \therefore \quad \mathrm{G}=\mathrm{s}(\mathrm{n}-1)...(i)
\end{aligned}\)
When same galvanometer is shunted with resistance ' \(\mathrm{s}_1\) '
\(\begin{aligned}
& \mathrm{s}_1=\mathrm{G}\left(\frac{1}{1-\mathrm{n}_1}\right) \\
& \mathrm{n}_1=\frac{\mathrm{G}+\mathrm{s}_1}{\mathrm{~s}_1}=\frac{\mathrm{s}(\mathrm{n}-1)+\mathrm{s}_1}{\mathrm{~s}_1}
\end{aligned}\)
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