Question

Resistance P, Q, S and R are arranged in a cyclic order to from a balanced Wheatstone’s network. The ratio of power consumed in the branches (P + Q) and (R + S) is

• A. 1 : 1
• B. R : P
• C. P2 : Q2
• D. P2 : R2

Answer 1

Answer: (B)

R : P

Answer 2

: For balanced wheatstone’s bridge,

$\frac{P}{Q} = \frac{R}{S}$

Power dissipation in resistance R with voltage V is

$V^{2}/R.$

$\therefore\frac{P_{p} + Q}{P_{R} + S} = \frac{R + S}{P + Q}$ ….(ii)

From equation (i),

$\frac{P}{Q} + 1 = \frac{R}{S} + 1 \Rightarrow \frac{P + Q}{Q} = \frac{R + S}{S}$ or $\frac{R + S}{P + Q} = \frac{S}{Q}$

Using (i), we get,

$\frac{R + S}{P + Q} = \frac{R}{P}$ $\therefore\frac{P_{P} + Q}{P_{R} + S} = \frac{R}{P}$