In the adjoining circuit,the potential differ | vedclass

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Question

(A) Let the total current entering the circuit be $I = 2\,A$. The circuit consists of two parallel branches.Branch $1$ contains a parallel combination

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the total current entering the circuit be $I = 2\,A$. The circuit consists of two parallel branches.Branch $1$ contains a parallel combination of $6\,\Omega$ and $3\,\Omega$ resistors in series with a $2\,\Omega$ resistor.The equivalent resistance of the parallel part ($6\,\Omega$ and $3\,\Omega$) is $R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\,\Omega$.Thus,the total resistance of the upper branch is $R_{upper} = 2\,\Omega + 2\,\Omega = 4\,\Omega$.The lower branch contains a $4\,\Omega$ resistor,so $R_{lower} = 4\,\Omega$.Since the two branches have equal resistance ($4\,\Omega$ each),the total current $I = 2\,A$ splits equally into both branches.Therefore,the current through the upper branch is $I_{upper} = 1\,A$.This current $I_{upper} = 1\,A$ flows through the parallel combination of $6\,\Omega$ and $3\,\Omega$ resistors.The potential difference across this parallel combination is $V_p = I_{upper} 	imes R_p = 1\,A 	imes 2\,\Omega = 2\,V$.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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