A `100 V` voltmeter having an internal resistance of `20 k Omega` is connected in series with a large resistance `R` across a `110 V` line. What is the magnitude of resistance `R` if the voltmeter reads `5 V`?

Explanation

When voltmeter connected in 110 volt line.
Current through the voltmeter `I=(110)/((20xx10^(3)+R))`
The potential difference across the voltmeter V=`IR_(v)`
`rArr=(110xx20xx10^(3))/((20x x10^(3)xxR))`
`rArr20xx10^(3)+R=440xx10^(3)`
`rArrR=420xx1-^(3) Omega`