Q. In the circuit shown in the figure, R = 55Ω, the equivalent resistance between the points P and Q is

(a) 30 Ω

(b) 35 Ω

(c) 55 Ω

(d) 25 Ω

Ans: (d)

Sol:

The resistance R/3 and R/2 in series

R’ = R/3 + R/2 = 5R/6

Now , 5R/6 & R are in parallel

$\displaystyle \frac{1}{R_{PQ}} = \frac{1}{R} + \frac{6}{5R} $

$ \displaystyle R_{PQ} = \frac{5R}{11} = \frac{5\times 55}{11} $

= 25 ohm