The equivalent resistance between AandB of the network shown in figure:

Question

Quantrex Academy · Accepted Answer

Correct answer: D. (8)/(3)R

The given circuit can be redrawn as shown below:

The equivalent resistance between AandB of the network shown in figure:

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

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

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

The equivalent resistance will therefore be:

${R}_{eq} = \dfrac{\left(4R\right)\left(8R\right)}{4R + 8R} = \dfrac{32{R}^{2}}{12R} = \dfrac{8}{3}R$

The equivalent resistance between AandB of the network shown in...