Question

The ratio of the amounts of heat developed in the four arms of a balanced Wheatstone bridge, when the arms have resistance $P=100\, \Omega ; Q=10\, \Omega ; R=300\, \Omega$ 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 Wheat stone bridge. Current through arms of resistances $P$ and $Q$ in series is
$I_{1}=\frac{1 \times 330}{330+110}=\frac{3}{4} I$ 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 $H_{P}: H_{Q}: H_{G}: H_{S}$
$\left.=\left(\frac{3}{4}\right)^{2} \times 100:\left(\frac{3}{4}\right)^{2} \times 10:\left(\frac{1}{4}\right)\right)^{2} \times 300:\left(\frac{1}{4} I\right)^{2} \times 30 $
$=30: 3: 10: 1$