In the given circuit,the internal resistance | vedclass

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(A) The resistors $R_3$ and $R_4$ are in series. The current $I_{upper}$ flowing through this branch is given by $I_{upper} = \frac{V_{R4}}{R_4} = \fr

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) The resistors $R_3$ and $R_4$ are in series. The current $I_{upper}$ flowing through this branch is given by $I_{upper} = \frac{V_{R4}}{R_4} = \frac{5\,V}{500\,\Omega} = 0.01\,A$.The voltage across the parallel combination of the upper branch $(R_3 + R_4)$ and $R_2$ is $V_p = I_{upper} 	imes (R_3 + R_4) = 0.01\,A 	imes (100\,\Omega + 500\,\Omega) = 0.01 	imes 600 = 6\,V$.The voltage across $R_1$ is $V_{R1} = V_{total} - V_p = 18\,V - 6\,V = 12\,V$.The current flowing through $R_1$ is $I_{total} = \frac{V_{R1}}{R_1} = \frac{12\,V}{400\,\Omega} = 0.03\,A$.Since $I_{total} = I_{upper} + I_{R2}$,the current through $R_2$ is $I_{R2} = I_{total} - I_{upper} = 0.03\,A - 0.01\,A = 0.02\,A$.Finally,the value of $R_2$ is $R_2 = \frac{V_p}{I_{R2}} = \frac{6\,V}{0.02\,A} = 300\,\Omega$.

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 $5\,V$,then the value of $R_2$ will be ........... $\Omega$.

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