Two batteries,one with an emf of 3,V and inte | vedclass

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(C) Let the current $I$ flow in the circuit in the anticlockwise direction.Applying Kirchhoff's Voltage Law $(KVL)$ to the entire loop:$15 + 3 = (1 +

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$3$

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(C) Let the current $I$ flow in the circuit in the anticlockwise direction.Applying Kirchhoff's Voltage Law $(KVL)$ to the entire loop:$15 + 3 = (1 + 2 + R) I$$18 = (3 + R) I$$I = \frac{18}{3 + R}$Given that the potential difference between points $a$ and $b$ is zero $(V_{ab} = 0)$:Starting from $a$ to $b$ through the $3\,V$ battery:$V_a - 3 + I(1) = V_b$$V_a - V_b = 3 - I = 0$$I = 3\,A$Substituting the value of $I$ in the loop equation:$3 = \frac{18}{3 + R}$$3 + R = 6$$R = 3\,\Omega$

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