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