Question 116 of 380

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

A$11\dfrac{2R}{3}$B$14R$C$21R$D$\dfrac{8}{3}R$

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Answer: (D) $\dfrac{8}{3}R$

The given circuit can be redrawn as shown below:

Since the circuit is in balanced wheatstone configuration, $\dfrac{R}{2R}=\dfrac{3R}{6R}$.
The $9\Omega$ 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:
${R}_{eq}=\dfrac{\left(4R\right)\left(8R\right)}{4R+8R}=\dfrac{32{R}^{2}}{12R}=\dfrac{8}{3}R$

JEE Main Physics practice question on Current Electricity (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.