If resistance of voltmeter is 10000$\Omega$ and resistance of galvanometer is 2$\Omega$, then find R when voltmeter reads 12V and galvanometer reads 0.1A.

Question

Dr. Abubacker · Accepted Answer

(a) The circuit diagram with current variation can be drawn as
Let current $I_{g}$flows through R and G.
For a voltmeter with full scale reading, we have
$\ \ \ \ \ \ \ V = I_{g}(G + R)$
$or\ \ \ R = \frac{V}{I_{g}} - G$
Given, $G = 2\Omega,\ V = 12volt,\ I_{g} = 0.1A$
$	herefore R = \frac{12}{0.1} - 2 = 120 - 2 = 118\Omega$

Q. If resistance of voltmeter is 10000$\Omega$ and resistance of galvanometer is 2$\Omega$, then find R when voltmeter reads 12V and galvanometer reads 0.1A.

Explanation

Reviewed by Dr. Abubacker

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