MHT CET · Physics · Current Electricity
When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is ' \(n\) ' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?
- A \(n\)
- B \(\frac {n+1}{n}\)
- C \(n+1\)
- D \(n-1\)
Answer & Solution
Correct Answer
(C) \(n+1\)
Step-by-step Solution
Detailed explanation
Series resistance is \(\mathrm{n}\) times of galvanometer resistance: \(\mathrm{R}_{\mathrm{s}}=\mathrm{nR}_{\mathrm{G}}\)
Relation between voltage and current:
\(\begin{array}{ll}
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{s}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{nR}_{\mathrm{G}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{IR}_{\mathrm{G}}(\mathrm{n}+1)
\end{array}\)
Relation between voltage and current:
\(\begin{array}{ll}
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{s}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{nR}_{\mathrm{G}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{IR}_{\mathrm{G}}(\mathrm{n}+1)
\end{array}\)
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