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A 100 V voltmeter of internal resistance...

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

Text Solution

Verified by Experts

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`
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