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(B) Given, the circuit diagram is shown in the figure. Let $I_1$ and $I_2$ be the loop currents in loop $(1)$ and loop $(2)$, respectively.By applying

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$\frac{8}{3} 	ext{ V}$

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(B) Given, the circuit diagram is shown in the figure. Let $I_1$ and $I_2$ be the loop currents in loop $(1)$ and loop $(2)$, respectively.By applying $KVL$ in loop $(1)$:$-5 + 10 I_1 + 10(I_1 - I_2) = 0$$20 I_1 - 10 I_2 = 5$  --- $(i)$By applying $KVL$ in loop $(2)$:$10 I_2 + 3 + 10(I_2 - I_1) = 0$$-10 I_1 + 20 I_2 = -3$  --- $(ii)$Multiplying Eq. $(i)$ by $2$ and adding to Eq. $(ii)$:$40 I_1 - 20 I_2 = 10$$-10 I_1 + 20 I_2 = -3$Adding these gives $30 I_1 = 7 \Rightarrow I_1 = \frac{7}{30} 	ext{ A}$.Substituting $I_1$ in Eq. $(i)$:$20(\frac{7}{30}) - 10 I_2 = 5$$\frac{14}{3} - 5 = 10 I_2$$10 I_2 = \frac{14 - 15}{3} = -\frac{1}{3} \Rightarrow I_2 = -\frac{1}{30} 	ext{ A}$.The current through the branch $AB$ is $I_{AB} = I_1 - I_2 = \frac{7}{30} - (-\frac{1}{30}) = \frac{8}{30} 	ext{ A}$.The voltage across terminals $AB$ is $V_{AB} = I_{AB} 	imes R_{AB} = \frac{8}{30} 	imes 10 = \frac{8}{3} 	ext{ V}$.

Calculate the voltage across $AB$ terminals in the given circuit.

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