Question

The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances $P = 100\, \Omega,\,Q = 10\, \Omega,\, R = 300\, \Omega$ and $S = 30\, \Omega$ respectively is

• A. 3 : 30 : 1 : 10
• B. 30 : 3 : 10 : 1
• C. 30 : 10 : 1 : 3
• D. 30 : 1 : 3 : 10

Answer 1

Answer: (B)

Let $i$ be the total current passing through balanced Wheatstone bridge. Current through arms of resistances $P$ and $Q$ in series is
$i_{1} =\frac{i \times 330}{330+110}$
$=\frac{3}{4} i$
and current through arms of resistances $R$ and $S$ in series is
$i_{2}=\frac{i \times 110}{330+110}=\frac{1}{4} i$
$\therefore $ Ratio of heat developed per sec
$H_{P}: H_{Q}: H_{R}: H_{S}$
$=\left(\frac{3}{4} i\right)^{2} \times 100:\left(\frac{3}{4} i\right)^{2} \times 10:\left(\frac{1}{4} i\right)^{2} \times 300$
$:\left(\frac{1}{4} i\right)^{2} \times 30$
$=30: 3: 10: 1$