The equivalent resistance and potential diffe | vedclass

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(A) First,we find the equivalent resistance between points $C$ and $D$. The three resistors of $6\,\Omega, 6\,\Omega$,and $3\,\Omega$ are connected in

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$4 \,\Omega, 8\, V$

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(A) First,we find the equivalent resistance between points $C$ and $D$. The three resistors of $6\,\Omega, 6\,\Omega$,and $3\,\Omega$ are connected in parallel.The equivalent resistance $R_{CD}$ is given by:$\frac{1}{R_{CD}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = \frac{1+1+2}{6} = \frac{4}{6} = \frac{2}{3}$$R_{CD} = 1.5\,\Omega$Now,the total resistance $R_{AB}$ between $A$ and $B$ is the series combination of $R_{CD}$ and the $2.5\,\Omega$ resistor:$R_{AB} = R_{CD} + 2.5\,\Omega = 1.5\,\Omega + 2.5\,\Omega = 4.0\,\Omega$Using Ohm's law,the potential difference between $A$ and $B$ is:$V_{AB} = I 	imes R_{AB} = 2\,A 	imes 4.0\,\Omega = 8\,V$Thus,the equivalent resistance is $4\,\Omega$ and the potential difference is $8\,V$.

The equivalent resistance and potential difference between $A$ and $B$ for the circuit shown are respectively:

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