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A 100 V voltmeter of internal resistance...

A `100 V` voltmeter of internal resistance `20 k Omega` in series with a high resistance `R` is connected to a `110 V` line. The voltmeter reads `5 V`, the value of `R` is

A

`420kOmega`

B

`42kOmega`

C

`4.2kOmega`

D

`420Omega`

Text Solution

Verified by Experts

The correct Answer is:
A

Potential difference across voltmeter `=5V`
`therefore` Current in circuit `=(5)/(20xx10^3)`
Potential difference across `R=105 V=I R`
`therefore R=(105)/(0.25xx10^-3)=420Omega`
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