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(D) The circuit consists of three parallel branches connected between points $P$ and $Q$. $1$. The left branch has resistors $R_A = 2 \Omega$ and $R_D

Vedclass · Accepted Answer

$2.88$

Vedclass · Answer

(D) The circuit consists of three parallel branches connected between points $P$ and $Q$. $1$. The left branch has resistors $R_A = 2 \Omega$ and $R_D = 6 \Omega$ in series. The resistance of this branch is $R_1 = 2 + 6 = 8 \Omega$. $2$. The middle branch has a resistor of $3 \Omega$. So, $R_2 = 3 \Omega$. $3$. The right branch has resistors $R_B = 4 \Omega$ and $R_C = 12 \Omega$ in series. The resistance of this branch is $R_3 = 4 + 12 = 16 \Omega$. Since these three branches are in parallel, the equivalent resistance $R_{PQ}$ is given by: $\frac{1}{R_{PQ}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{8} + \frac{1}{3} + \frac{1}{16}$. Taking the least common multiple $(LCM)$ of $8, 3, 16$, which is $48$: $\frac{1}{R_{PQ}} = \frac{6 + 16 + 3}{48} = \frac{25}{48} \Omega^{-1}$. Therefore, $R_{PQ} = \frac{48}{25} \Omega = 1.92 \Omega$. The total current $I = 1.5 	ext{ A}$ flows through this equivalent resistance. The potential difference $V_{PQ}$ is: $V_{PQ} = I \cdot R_{PQ} = 1.5 	imes 1.92 = 2.88 	ext{ V}$.

The potential difference between the points $P$ and $Q$ in the circuit shown is: (in $\text{ V}$)

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