A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivity of 20 μ A / d i v i s i o n . It is to be converted to voltmeter with three ranges, of 0 - 2 V , 0 - 10 V and 0 - 2

Question

A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivity of 20 μ A / d i v i s i o n . It is to be converted to voltmeter with three ranges, of 0 - 2 V , 0 - 10 V and 0 - 20 V . The appropriate circuit to do so is:

TestHub · Accepted Answer

Given, the resistance of the galvanometer $G = 100 Ω$
No. of divisions $= 50$
Current Sensitivity $= 20 μ A / d i v i s i o n$

The maximum current through galvanometer will be
$I_{m a x} = 50 \times 20 \times 10^{- 6} = 10^{- 3} A$
For $V_{1} = 2 V$

using Ohm's law $V_{1} = I_{m a x} (G + R_{1})$
$2 = 10^{- 3} \times (100 + R_{1})$
$∴ R_{1} = 1900 Ω$

For $V_{2} = 10 V$

using Ohm's law $V_{2} = I_{m a x} (G + R_{1} + R_{2})$

$10 = 10^{- 3} \times (100 + 1900 + R_{2})$
$∴ R_{2} = 8000 Ω$

For $V_{3} = 20 V$

using Ohm's law $V_{3} = I_{m a x} (G + R_{1} + R_{2} + R_{3})$
$20 = 10^{- 3} (100 + 1900 + 8000 + R_{3})$
$∴ R_{3} = 10 \times 10^{3} Ω$

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