Question:
A galvanometer having a coil resistance \(100\ \Omega\) gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?
A galvanometer having a coil resistance \(100\ \Omega\) gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?
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A series resistance \(R\) makes total resistance \(V/I_g\). So \(R = \frac{V}{I_g} - G\).
Updated On: Nov 5, 2026
- \(8.9\ \mathrm{k\Omega}\)
- \(10\ \mathrm{k\Omega}\)
- \(9.9\ \mathrm{k\Omega}\)
- \(7.9\ \mathrm{k\Omega}\)
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The Correct Option is C
Solution and Explanation
To make a voltmeter, connect a high resistance \(R\) in series with the galvanometer. For full-scale deflection, \(I_g = 1\) mA flows when \(V = 10\) V is applied:
\(I_g (G + R) = V \Rightarrow R = \frac{V}{I_g} - G = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega = 9.9\ \mathrm{k\Omega}\).
Answer: (3) \(9.9\ \mathrm{k\Omega}\)
\(I_g (G + R) = V \Rightarrow R = \frac{V}{I_g} - G = \frac{10}{10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega = 9.9\ \mathrm{k\Omega}\).
Answer: (3) \(9.9\ \mathrm{k\Omega}\)
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