Question

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

Answer 1

Given:

Resistence of galvanometer = 40Ω

Current = 10mA

Voltage = 50 V

To Find:

The resistance to be connected in the series with the galvanometer is

Solution:

Let the resistance of a resistor be =  R  

R is connected in series with galvanometer resistance to convert it in a voltmeter.

Thus,

V = Ig  (G+R)

= 50 = 10 × 10-³  (40+R)

5000 = 40+R                

R = 5000 - 40

R = 4960

Answer: The resistance to be connected in the series with the galvanometer is 4960Ω