A cell of emf 1.4V and internal resistance `2 Omega` is connected in series with a resistance of `100 Omega` and an ammeter. The resistance of the ammeter is `4/3 Omega`.To measure the potential difference between the two ends of the resistance a voltmeter is connected IF the reading of the ammter is 0.02 A.what is the resistance of the voltmeter?

Explanation

Lost volt of the cell=Ir=0.02 `times `2=0.04 V
Potential difference between the two ends of the ammter
`=0.02 times 4/3=0.08/3 V`
So, `V_(CD)=1.4-0.04-0.08/3=(4.2-0.12-0.08)/3=4/3V`

Therefore, current through `100 Omega` resistance
`V_(CD)/100=4/300A`
So, current through the voltmeter
`=0.02-4/300=2/300A`
Therefore, resistance of the voltmeter
`=V_(CD)/(2/300)=4/3times300/2=200 Omega`