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Question

(A) The circuit is a Wheatstone bridge. Let the resistors be $P = 3 \, \Omega$,$Q = 4 \, \Omega$,$R = 6 \, \Omega$,and $S = 8 \, \Omega$. The central

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The circuit is a Wheatstone bridge. Let the resistors be $P = 3 \, \Omega$,$Q = 4 \, \Omega$,$R = 6 \, \Omega$,and $S = 8 \, \Omega$. The central resistor is $G = 7 \, \Omega$.
Check the ratio of the arms: $\frac{P}{R} = \frac{3}{6} = 0.5$ and $\frac{Q}{S} = \frac{4}{8} = 0.5$.
Since $\frac{P}{R} = \frac{Q}{S}$,the Wheatstone bridge is balanced.
In a balanced Wheatstone bridge,no current flows through the central resistor $G$. Therefore,it can be removed from the circuit.
After removing the $7 \, \Omega$ resistor,the circuit consists of two parallel branches:
Branch $1$ (upper): $3 \, \Omega$ and $4 \, \Omega$ in series,so $R_1 = 3 + 4 = 7 \, \Omega$.
Branch $2$ (lower): $6 \, \Omega$ and $8 \, \Omega$ in series,so $R_2 = 6 + 8 = 14 \, \Omega$.
The equivalent resistance $R_{eq}$ between $A$ and $B$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{7} + \frac{1}{14}$.
$\frac{1}{R_{eq}} = \frac{2 + 1}{14} = \frac{3}{14}$.
Thus,$R_{eq} = \frac{14}{3} \, \Omega$.

Calculate the net resistance of the circuit between $A$ and $B$.

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