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Q. In a typical Wheatstone's network the resistance in cyclic order are P = 10 ohm, Q = 5 ohm, S = 4 ohm and R = 4 ohm. For the bridge to balance

KCETKCET 2000Current Electricity

Solution:

and 4 Here $\frac{R}{S} = \frac{4}{4}$ = 1 and $\frac{P}{Q} = \frac{10}{5}$
If 5$\Omega$ resistor is added in series with Q then $\frac{P}{Q} = \frac{10}{10} = 1$