A galvanometer of resistance 100Ω 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

Question

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

The maximum current through galvanometer will be
$I_{ma x} = 50 \times 20 \times 1 0^{- 6} = 1 0^{- 3} A$
For $V_{1} = 2 V$

using Ohm's law $V_{1} = I_{ma x} (G + R_{1})$
$2 = 1 0^{- 3} \times (100 + R_{1})$
$∴ R_{1} = 1900Ω$

For $V_{2} = 10 V$

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

$10 = 1 0^{- 3} \times (100 + 1900 + R_{2})$
$∴ R_{2} = 8000Ω$

For $V_{3} = 20 V$

using Ohm's law $V_{3} = I_{ma x} (G + R_{1} + R_{2} + R_{3})$
$20 = 1 0^{- 3} (100 + 1900 + 8000 + R_{3})$
$∴ R_{3} = 10 \times 1 0^{3} Ω$

Physics - Current Electricity Question with Solution | TestHub

Options:

Doubts & Discussion