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(A) The circuit consists of a parallel combination of a branch containing ($6\, \Omega$ and $3\, \Omega$ in parallel,followed by $2\, \Omega$ in serie

Vedclass · Accepted Answer

$2$

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(A) The circuit consists of a parallel combination of a branch containing ($6\, \Omega$ and $3\, \Omega$ in parallel,followed by $2\, \Omega$ in series) and a branch containing $4\, \Omega$.First,calculate the equivalent resistance of the upper branch. The $6\, \Omega$ and $3\, \Omega$ resistors are in parallel,so their equivalent resistance $R_p$ is given by $R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\, \Omega$.This $2\, \Omega$ is in series with the other $2\, \Omega$ resistor,so the total resistance of the upper branch is $R_{upper} = 2\, \Omega + 2\, \Omega = 4\, \Omega$.The lower branch also has a resistance of $4\, \Omega$.Since both branches have equal resistance $(4\, \Omega)$,the total current of $2\, A$ splits equally between them. Thus,$1\, A$ flows through the upper branch.The potential difference across the parallel combination of $6\, \Omega$ and $3\, \Omega$ is $V = I_{upper} 	imes R_p = 1\, A 	imes 2\, \Omega = 2\, V$.Since the $6\, \Omega$ and $3\, \Omega$ resistors are in parallel,the potential difference across each is the same,which is $2\, V$.

In the adjoining circuit,the potential difference across $3\, \Omega$ is ............... $V$.

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