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A voltmeter reads the potential differen...

A voltmeter reads the potential difference across the terminals of an old battery as `1.40 V`, while a potentiometer reads its voltage to be `1.55 V`. The voltmeter resistance is `280 Omega`.

A

`60Omega`

B

`45Omega`

C

`35Omega`

D

`30Omega`

Text Solution

Verified by Experts

The correct Answer is:
4
Let `epsilon` be emf and r be internal resistance of the cell when tno current drawn from the cell, the potential difference b etween its terminals is the emf of the cell.

It is this quantity which the potentionmeter measures.
Therefore, emf of the cell `epsilon=1.55V`
When the `280Omega` voltmeter is connected ac ross the cell as shown in adjacent figure, current flows from the cell.
The potential difference between its terminals is just the voltmeter reading. 140V.
`terefore 1xx280=1.40`
`1=(1.4)/(280)=5xx10^(-3)A`
Also `I=(epsilon)/(280+r)=(1.55)/(28+r)`
`5xx10^(-3)=(1.55)/(28+r)`
`1.4+5xx10^(-3)r=1.55`
`5xx10^(-3)r=0.15`
`r=(0.15)/(5xx10^(-3)=30Omega`
According to m aximum power theoram.
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