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(D) The circuit consists of three parallel branches connected between points $P$ and $Q$.$1$. The first branch contains resistors $R_A = 2\, \Omega$ a

Vedclass · Accepted Answer

$2.88$

Vedclass · Answer

(D) The circuit consists of three parallel branches connected between points $P$ and $Q$.$1$. The first branch contains 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 second branch contains a single resistor of $3\, \Omega$. So,$R_2 = 3\, \Omega$.$3$. The third branch contains resistors $R_B = 4\, \Omega$ and $R_C = 12\, \Omega$ in series. The resistance of this branch is $R_3 = 4 + 12 = 16\, \Omega$.Now,calculate the equivalent resistance $R_{PQ}$ of these three parallel branches:$\frac{1}{R_{PQ}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{8} + \frac{1}{3} + \frac{1}{16}$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 entering the junction $P$ is $i = 1.5\, A$. The potential difference $V_{PQ}$ is given by Ohm's Law:$V_{PQ} = i 	imes R_{PQ} = 1.5 	imes \frac{48}{25} = 1.5 	imes 1.92 = 2.88\, V$.

The potential difference between the points $P$ and $Q$ in the electric circuit shown is ................. $V$.

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