Q. In the given circuit the internal resistance of the 18 V cell is negligible. If $R_1 = 400 \; \Omega , R_3 = 100 \Omega$ and $R_4 = 500 \; \Omega $ and the reading of an ideal voltmeter across $R_4$ is 5V, then the value $R_2$ will be :

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Solution:

$V_{4} = 5V $
$ i_{1} = \frac{V_{4}}{R_{4}} = 0.01 A$
$ V_{3} = i_{1} R_{3} = 1V$
$ V_{3} +V_{4} = 6V = V_{2} $
$ V_{1} + V_{3} + V_{4} = 18 V $
$ V_{1} = 12V $
$ i = \frac{V_{1} }{R_{1}} = 0.03 Amp .$
$ i_{2} = 0.02 Amp V_{2} = 6V $
$ R_{2} = \frac{V_{2} }{i_{2}} = \frac{6}{0.02} = 300 \Omega$

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