The potential difference across the 8 Omega r | vedclass

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(A) The circuit can be simplified by analyzing the parallel combinations of resistors.First,consider the lower parallel branch containing $24\ \Omega$

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$160$

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(A) The circuit can be simplified by analyzing the parallel combinations of resistors.First,consider the lower parallel branch containing $24\ \Omega$ and $8\ \Omega$ resistors. The equivalent resistance $R_{AB}$ between points $A$ and $B$ is given by:$R_{AB} = \frac{24 	imes 8}{24 + 8} = \frac{192}{32} = 6\ \Omega$.Given that the potential difference across the $8\ \Omega$ resistor is $48\ V$,the current $i$ flowing through the branch $AB$ is:$i = \frac{48\ V}{6\ \Omega} = 8\ A$.Next,consider the upper parallel branch containing $20\ \Omega$,$30\ \Omega$,and $60\ \Omega$ resistors. The equivalent resistance $R_{upper}$ is:$\frac{1}{R_{upper}} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3+2+1}{60} = \frac{6}{60} = \frac{1}{10} \Rightarrow R_{upper} = 10\ \Omega$.The total resistance between $X$ and $Y$ is the sum of the series components:$R_{XY} = 3\ \Omega + R_{upper} + R_{AB} + 1\ \Omega = 3 + 10 + 6 + 1 = 20\ \Omega$.The potential difference across $X$ and $Y$ is $V_{XY} = i 	imes R_{XY} = 8\ A 	imes 20\ \Omega = 160\ V$.

The potential difference across the $8\ \Omega$ resistance is $48\ V$ as shown in the figure. The value of the potential difference across points $X$ and $Y$ will be ............... $V$.

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