The maximum current that can be measured by a galvanometer of resistance 40 Ω is 10 mA. It is converted into a voltmeter that can read upto 50 V. The resistance to be connected in series with the galvanometer is .......... (in ohm)

Question

The maximum current that can be measured by a galvanometer of resistance 40 Ω is 10 mA. It is converted into a voltmeter that can read upto 50 V. The resistance to be connected in series with the galvanometer is .......... (in ohm) (A) 5040 (B) 4960 (C) 2010 (D) 4050

Tardigrade Admin · Accepted Answer

To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is R = (V/Ig) - G = (50/10 × 10-3) - 40 = 5000 - 40 = 4960Ω

Q. The maximum current that can be measured by a galvanometer of resistance $40 Ω$ is $10 m A$ . It is converted into a voltmeter that can read upto $50 V$ . The resistance to be connected in series with the galvanometer is .......... (in ohm)

5040

4960

2010

4050

To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is
$R = \frac{V}{I _{g}} - G$
$= \frac{50}{10 \times 1 0 ^{- 3}} - 40$
$= 5000 - 40$
$= 4960Ω$

Q. The maximum current that can be measured by a galvanometer of resistance 40Ω is 10mA. It is converted into a voltmeter that can read upto 50V. The resistance to be connected in series with the galvanometer is .......... (in ohm)