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A galvanometer has a full scale deflection current of 10 mA and it has a small internal resistance of $20 Ω .$ The resistance to be connected in series with it to convert it into a voltmeter of range 0 to 10 V is $R_{1}$ . The resistance to be connected in series with it to convert it into a voltmeter of range 0 to 20 V is $R_{2}$ . The ratio $R_{1} / R_{2}$ is nearly equal to

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1 : 4

2 : 1

1 : 2

4 : 1

answer is B.

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Detailed Solution

$i_{g} = 10 m A$

$G = 20 Ω$

$R_{S} = \frac{V}{i_{g}} - G$

$R_{1} = \frac{10}{10 \times 10^{- 3}} - 20 = 980$

$R_{2} = \frac{20}{10 \times 10^{- 3}} - 20 = 1980$

$\frac{R_{1}}{R_{2}} = \frac{1}{2}$

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