A battery of e.m.f. 10, V is connected to a r | vedclass

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(B) $1$. First,calculate the equivalent resistance of the circuit. The two parallel branches each have a resistance of $(1 + 3) = 4\,\Omega$. The equi

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$2$

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(B) $1$. First,calculate the equivalent resistance of the circuit. The two parallel branches each have a resistance of $(1 + 3) = 4\,\Omega$. The equivalent resistance of these two parallel branches is $R_p = \frac{4 	imes 4}{4 + 4} = 2\,\Omega$.$2$. The total resistance of the circuit is $R_{eq} = 3\,\Omega + 2\,\Omega = 5\,\Omega$.$3$. The total current $i$ flowing from the battery is $i = \frac{V}{R_{eq}} = \frac{10}{5} = 2\,A$.$4$. Since the two parallel branches have equal resistance,the current splits equally,so $i_1 = i_2 = \frac{i}{2} = 1\,A$.$5$. Let $C$ be the junction point where the current splits. The potential at $A$ relative to $C$ is $V_C - V_A = i_1 	imes 1\,\Omega = 1 	imes 1 = 1\,V$.$6$. The potential at $B$ relative to $C$ is $V_C - V_B = i_2 	imes 3\,\Omega = 1 	imes 3 = 3\,V$.$7$. Subtracting the two equations: $(V_C - V_A) - (V_C - V_B) = 1 - 3$,which simplifies to $V_B - V_A = -2\,V$,or $V_A - V_B = 2\,V$.

$A$ battery of $e.m.f.$ $10\, V$ is connected to a resistance network as shown in the figure. The potential difference ${V_A} - {V_B}$ between the points $A$ and $B$ is .................... $V$.

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