Q. 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
Explanation
(b) Let $I$ be the total current passing through balanced wheat
stone bridge. Current through arms of resistances $P$ and
$Q$ in series is
$I_{1} = \frac{I \times 300}{330 + 110} = \frac{3}{4}I$ 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
$H_{P}\ :H_{Q}\ :H_{R}\ :H_{S}$
$= \left( \ \frac{3}{4}I \right)^{2} \times 100\ :\left( \ \frac{3}{4}I \right)^{2} \times 10\ :\left( \ \frac{1}{4}I \right)^{2} \times 300\ :\left( \ \ \frac{1}{4}I \right)^{2} \times 30$
$= 30\ :3\ :10\ :1$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.