Question

Five equal resistances each of value R are connected to form a network as shown in figure. The equivalent resistance of the network between the points A and B is

• A. $T_{2}$
• B. 2 R
• C. $T_{1} > T_{2}$
• D. $T_{1} < T_{2}$

Answer 1

Answer: (C)

$T_{1} > T_{2}$

Answer 2

: The given circuit is redrawn as shown in the figure.

The resistances connected between BC and CD are in series. Therefore its equivalent resistance is 2R. The resistance 2R and the resistance connected between BD are in parallel. Let its equivalent resistance be $R_{1}$

$\therefore\frac{1}{R_{1}} = \frac{1}{2R} + \frac{1}{R}$ or $R_{1} = \frac{2}{3}R$

The resistance $R_{1}$and resistance R connected between AD are in series. Let its equivalent resistance be $R_{2}$.

$\therefore R_{2} = R + \frac{2}{3}R = \frac{5}{3}R$

The resistance $R_{2}$and resistance R connected between AB are in parallel. Hence the equivalent resistance between AB is $R_{eq}$

$\therefore\frac{1}{R_{eq}} = \frac{1}{\frac{5}{3}R} + \frac{1}{R}$ or $R_{eq} = \frac{5R}{8}$