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

As shown in Fig. `S6.3`, the voltmeter
is in series with `R`, so `V = V_(1) + V_(R)`, i.e.,
`110 = 5 + V_(R)`
i.e., `V_(R) = 110 -5 = 105 V`
And as in series, potential divides in proportion to resistance, i.e.,
`(V_(R))/(V_(1)) = (R )/(R_(1))` or `(105)/(5) = (R )/(20 k Omega)` or `R = 420 k Omega`