Question

A high resistance voltmeter connected across the terminals of abattery reads 15v.when an ammeter is included in the circuit voltmeter reads 9v and ammeter reads 1.5



a.find internal resistance

Answer 1

#ANSWER #

Let, ‘r’ be the internal resistance of the battery and ‘R’ be the resistance of the ammeter, r’ = resistance of the connecting wires.

When battery is connected with the ammeter. The current through the circuit is I = 1.5 A.

Since resistance of the voltmeter is very high current through it is negligible thus E = 15 V = emf of the cell

We have

V = potential drop at ammeter

E = V + Ir + Ir’

But now I = 1.5

But V = IR that is

=> 1.5(R +r+r’)= 15

=> R+r+r’ = 10------------------1.

Potential difference across the cell will be due to the internal resistance r of the cell

But Ir = 9 (given)

 1.5r= 9

=>r = 6    -------------------2

Putting eqn. 2 in 1

R+6+r’=10

Thus,

Internal resistance = 6 ohm

Resistance of ammeter + connecting wires = 4 ohm.


Thanx

Answer 2

Answer:internal resistance=4 ohm

Resistance of ammeter & connecting wire=6 ohm