The potential difference across the 8 Omega | vedclass

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(A) First,simplify the circuit. The parallel combination of $20 \ \Omega$,$30 \ \Omega$,and $60 \ \Omega$ resistors is: $\frac{1}{R_p} = \frac{1}{20}

Vedclass · Accepted Answer

$160$

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(A) First,simplify the circuit. The parallel combination of $20 \ \Omega$,$30 \ \Omega$,and $60 \ \Omega$ resistors is: $\frac{1}{R_p} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3+2+1}{60} = \frac{6}{60} = \frac{1}{10} \implies R_p = 10 \ \Omega$.Similarly,the parallel combination of $24 \ \Omega$ and $8 \ \Omega$ resistors is: $R_p' = \frac{24 	imes 8}{24 + 8} = \frac{192}{32} = 6 \ \Omega$.The circuit simplifies to a series arrangement of $3 \ \Omega$,$10 \ \Omega$,$6 \ \Omega$,and $1 \ \Omega$ resistors.Given that the potential difference across the $6 \ \Omega$ equivalent resistance is $48 \ V$,the current $I$ flowing through this series circuit is $I = \frac{V}{R} = \frac{48}{6} = 8 \ A$.The total resistance of the circuit is $R_{eq} = 3 + 10 + 6 + 1 = 20 \ \Omega$.Therefore,the potential difference across points $X$ and $Y$ is $V_{XY} = I 	imes R_{eq} = 8 \ A 	imes 20 \ \Omega = 160 \ V$.

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: (in $V$)

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