50,Omega and 100,Omega resistors are connecte | vedclass

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(C) First,calculate the equivalent resistance of the parallel combination of the $100\,\Omega$ resistor and the $100\,\Omega$ voltmeter: $R_p = \frac{

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$1.2$

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(C) First,calculate the equivalent resistance of the parallel combination of the $100\,\Omega$ resistor and the $100\,\Omega$ voltmeter: $R_p = \frac{100 	imes 100}{100 + 100} = 50\,\Omega$.Now,the total resistance of the circuit is $R_{eq} = 50\,\Omega$ (series resistor) $+ 50\,\Omega$ (parallel combination) $= 100\,\Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{2.4\,V}{100\,\Omega} = 0.024\,A$.The voltmeter reading is the potential difference across the parallel combination: $V_{voltmeter} = I 	imes R_p = 0.024\,A 	imes 50\,\Omega = 1.2\,V$.

$50\,\Omega$ and $100\,\Omega$ resistors are connected in series. This combination is connected to a battery of $2.4\,V$. When a voltmeter of $100\,\Omega$ resistance is connected across the $100\,\Omega$ resistor,the reading of the voltmeter will be ............. $V$.

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