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(C) The circuit consists of a $60\,V$ source connected to a series combination of a $300\,\Omega$ resistor and a $400\,\Omega$ resistor. Let the voltm

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$22.5$

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(C) The circuit consists of a $60\,V$ source connected to a series combination of a $300\,\Omega$ resistor and a $400\,\Omega$ resistor. Let the voltmeter have an internal resistance $R_v$.When the voltmeter is connected across the $400\,\Omega$ resistor,the equivalent resistance of the parallel combination is $R_p = \frac{400 R_v}{400 + R_v}$.The total resistance of the circuit is $R_{eq} = 300 + R_p$.The voltage across the parallel combination is given as $30\,V$. Thus,the voltage across the $300\,\Omega$ resistor is $60 - 30 = 30\,V$.Since the voltage across both parts is equal $(30\,V)$,the resistance of the parallel combination must be equal to the resistance of the $300\,\Omega$ resistor.Therefore,$R_p = 300\,\Omega$.Now,when the voltmeter is connected across the $300\,\Omega$ resistor,the new equivalent resistance of that part is $R'_p = \frac{300 R_v}{300 + R_v}$.Since $R_p = \frac{400 R_v}{400 + R_v} = 300$,we find $400 R_v = 120000 + 300 R_v$,so $100 R_v = 120000$,which means $R_v = 1200\,\Omega$.Now,$R'_p = \frac{300 	imes 1200}{300 + 1200} = \frac{360000}{1500} = 240\,\Omega$.The total resistance of the circuit is now $R'_{eq} = 400 + 240 = 640\,\Omega$.The total current is $I = \frac{60}{640} = \frac{6}{64} = \frac{3}{32}\,A$.The reading of the voltmeter is $V' = I 	imes R'_p = \frac{3}{32} 	imes 240 = 3 	imes 7.5 = 22.5\,V$.

In the given circuit,the reading of the voltmeter connected across the $400\,\Omega$ resistor is $30\,V$. If it is connected across the $300\,\Omega$ resistor,what will be its reading in $V$?

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