The equivalent resistance between A a n d B of the network shown in figure:

Question

TestHub · Accepted Answer

The given circuit can be redrawn as shown below:

Since the circuit is in balanced wheatstone configuration, $\frac{R}{2 R} = \frac{3 R}{6 R}$ .

The $9 Ω$ resistor can be omitted. Now, the circuit has two branches having two resistors connected in series.

Therefore, the resistance of upper branch is $R + 3 R = 4 R$ and lower branch is $2 R + 6 R = 8 R$ .

The equivalent resistance will therefore be:

$R_{e q} = \frac{(4 R) (8 R)}{4 R + 8 R} = \frac{32 R^{2}}{12 R} = \frac{8}{3} R$

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