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Answer: A resistance of 49900 Ω (49.9 kΩ) should be connected in series with the galvanometer.

Given \(G = 100\ \Omega\), \(I_g = 0.2\) mA \(= 2\times10^{-4}\) A, \(V = 10\) V.

To convert it into a voltmeter, a high resistance \(X\) is put in series:

\[V = I_g(G + X) \Rightarrow X = \frac{V}{I_g} - G = \frac{10}{2\times10^{-4}} - 100 = 50000 - 100 = 49900\ \Omega\]
Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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