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 $H_P:H_Q:H_G:H_S$
${={\left(\frac{3}{4}\right)}^2\times 100:{\left(\frac{3}{4}\right)}^2\times 10:\left(\frac{1}{4}\right))}^2\times 300:{\left(\frac{1}{4}I\right)}^2\times 30$
$=30:3:10:1$