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(D) The given circuit is a Wheatstone bridge. Let the nodes be labeled such that the resistors $R$,$3R$,$2R$,and $6R$ form the four arms of the bridge

Vedclass · Accepted Answer

$\frac{8R}{3}$

Vedclass · Answer

(D) The given circuit is a Wheatstone bridge. Let the nodes be labeled such that the resistors $R$,$3R$,$2R$,and $6R$ form the four arms of the bridge.Checking the condition for a balanced Wheatstone bridge: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.Here,$\frac{R}{2R} = \frac{1}{2}$ and $\frac{3R}{6R} = \frac{1}{2}$.Since the ratios are equal,the bridge is balanced,and the central resistor $(9R)$ carries no current.Thus,the circuit simplifies to two parallel branches: one with $(R + 3R) = 4R$ and the other with $(2R + 6R) = 8R$.The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{8R} = \frac{2+1}{8R} = \frac{3}{8R}$.Therefore,$R_{eq} = \frac{8R}{3}$.

Find the equivalent resistance between points $A$ and $B$ for the network shown in the figure.

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