A 100 V voltmeter having a resistance of 20 | vedclass

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(A) The voltmeter and the resistance $R$ are connected in series across a $110 \ V$ supply.Let $R_v = 20 \ k\Omega = 20 	imes 10^3 \ \Omega$ be the r

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$420 	imes 10^3 \ \Omega$

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(A) The voltmeter and the resistance $R$ are connected in series across a $110 \ V$ supply.Let $R_v = 20 \ k\Omega = 20 	imes 10^3 \ \Omega$ be the resistance of the voltmeter.The total resistance of the circuit is $R_{total} = R_v + R$.The current $I$ flowing through the circuit is given by $I = \frac{V_{total}}{R_v + R} = \frac{110}{20 	imes 10^3 + R}$.The voltage across the voltmeter is given as $V_v = 5 \ V$.Using Ohm's law for the voltmeter,$V_v = I 	imes R_v$.Substituting the values: $5 = \left( \frac{110}{20 	imes 10^3 + R} ight) 	imes 20 	imes 10^3$.$5(20 	imes 10^3 + R) = 110 	imes 20 	imes 10^3$.$100 	imes 10^3 + 5R = 2200 	imes 10^3$.$5R = 2100 	imes 10^3$.$R = 420 	imes 10^3 \ \Omega$.

$A$ $100 \ V$ voltmeter having a resistance of $20 \ k\Omega$ is connected in series with a very high resistance $R$. When it is connected to a $110 \ V$ line,it reads $5 \ V$. What is the value of $R$?

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