In the electrical circuit shown in the figure | vedclass

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Question

(B) Let the total current from the battery be $I$ and the current through the right loop be $I_1$. Applying Kirchhoff's voltage law to the right loop

Vedclass · Accepted Answer

$0.5$

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(B) Let the total current from the battery be $I$ and the current through the right loop be $I_1$. Applying Kirchhoff's voltage law to the right loop (loop $BCDEB$): $2 I_1 + 4 I_1 + 2 I_1 - 8(I - I_1) = 0$ $8 I_1 - 8 I + 8 I_1 = 0$ $16 I_1 = 8 I \Rightarrow I_1 = 0.5 I$. Now, applying Kirchhoff's voltage law to the left loop (loop $ABEF$): $9 - 3 I - 8(I - I_1) - 2 I = 0$ $9 - 5 I - 8(I - 0.5 I) = 0$ $9 - 5 I - 8(0.5 I) = 0$ $9 - 5 I - 4 I = 0$ $9 I = 9 \Rightarrow I = 1 A$. Therefore, the current through the $4 \Omega$ resistor is $I_1 = 0.5 	imes 1 A = 0.5 A$.

In the electrical circuit shown in the figure, the current through the $4 \Omega$ resistor is: (in $A$)

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