Question

A galvanometer of coil resistance 1 is converted into voltmeter by using a resistance of 5 in series and same galvanometer is converted intoammeter by using a shunt of 1. Now ammeter and voltmeter connected in circuit as shown, find the reading of voltmeter and ammeter.​

Answer 1

Answer:

when connected as ammeter effective resistance of circuitR=r+99//1+2=r+0.99+2

current in circuit =

R

12

=3⇒

3

12

=R⇒4=r+0.99+2⇒r=1.01Ω

consider the voltmeter circuit,

current in circuit =

r+2

12

=4A

voltage across 2 ohm resistance =

2+1

2

12=8V

this is 4/5th of full scale reading.

Therefore full scale reading of voltmeter=

4

5

8=10V

current passing through the voltmeter=

200+2

2

4≈0.04A

this is 4/5th of maximum current allowed in the galvanometer when used as voltmeter.

therefore ,

maximum current allowed in galvanometer when used as voltmeter=

4

5

0.04=0.05A

therefore full scale reading of galvanometer is 0.05A . Therefore x=5