In the circuit shown,the current through R_2 | vedclass

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(B) Given that the current through $R_2$ is zero,the potential at the top node (let's call it $V_A$) must be equal to the potential at the bottom node

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$5$

Vedclass · Answer

(B) Given that the current through $R_2$ is zero,the potential at the top node (let's call it $V_A$) must be equal to the potential at the bottom node (let's call it $V_B$).Looking at the right side of the circuit,the node connected to the $5\,V$ battery and $R_4$ is grounded $(0\,V)$.Since the current through $R_2$ is zero,the potential at the junction between $R_1, R_2$ and the $5\,V$ battery is $-5\,V$ (because the $5\,V$ battery is connected to $0\,V$ ground).Similarly,the potential at the junction between $R_2, R_3$ and $R_4$ is $-5\,V$ (because the current through $R_4$ flows from $0\,V$ to $-5\,V$ is not possible,rather the potential drop across $R_4$ must be consistent with the circuit).Actually,the potential at the node between $R_3$ and $R_4$ is $-5\,V$ because the current through $R_4$ is $I = \frac{0 - (-5)}{2} = 2.5\,A$.The potential at the left side of $R_3$ is $15\,V$ relative to the ground.Therefore,the current through $R_3$ is $I = \frac{15 - (-5)}{4} = \frac{20}{4} = 5\,A$.

In the circuit shown,the current through $R_2$ is zero. If $R_4 = 2\,\Omega$ and $R_3 = 4\,\Omega$,the current through $R_3$ will be ................. $A$.

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