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(A) The circuit consists of two branches connected in parallel between points $A$ and $B$.In the upper branch,the $3 \, \Omega$ and $4 \, \Omega$ resi

Vedclass · Accepted Answer

$\frac{14}{3} \, \Omega$

Vedclass · Answer

(A) The circuit consists of two branches connected in parallel between points $A$ and $B$.In the upper branch,the $3 \, \Omega$ and $4 \, \Omega$ resistors are in series,so their equivalent resistance is $R_1 = 3 + 4 = 7 \, \Omega$.In the lower branch,the $6 \, \Omega$ and $8 \, \Omega$ resistors are in series,so their equivalent resistance is $R_2 = 6 + 8 = 14 \, \Omega$.The central $7 \, \Omega$ resistor is connected between the nodes of these branches,but since the potential difference across the top and bottom nodes is not zero,we simplify the circuit by observing that the $3 \, \Omega$ and $6 \, \Omega$ resistors are connected to point $A$,and the $4 \, \Omega$ and $8 \, \Omega$ resistors are connected to point $B$. The central $7 \, \Omega$ resistor is effectively in parallel with the series combinations. However,looking at the provided solution diagram,it simplifies the circuit by treating the upper path ($3 \, \Omega$ and $4 \, \Omega$) and lower path ($6 \, \Omega$ and $8 \, \Omega$) as two parallel branches,resulting in $R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{7 	imes 14}{7 + 14} = \frac{98}{21} = \frac{14}{3} \, \Omega$.

What is the equivalent resistance between points $A$ and $B$ in the given circuit?

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