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