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A 100 V voltmeter of internal resistance...

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

A

`210 k Omega `

B

` 315 k Omega `

C

` 420 k Omega `

D

`4440 k Omega `

Text Solution

Verified by Experts

The correct Answer is:
C
The circuit is shown below

Potential difference across voltmeter = 5 V
` :.` Current in the circuit, ` i = (5)/( 20 xx 10^(3))`
`= 0 . 25 xx 10 ^(-3) A `
Voltage across R,
`V_(1) = 110 - 5 = 105 V `
Hence, `R = (V_(1))/(i) = ( 105)/( 0 . 25 xx 10^(-3))` ` = 420 k Omega `
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