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

Updated On: Apr 15, 2024
  • 3 : 30 : 1 : 10
  • 30 : 3 : 10 : 1
  • 30 : 10 : 1 : 3
  • 30 : 1 : 3 : 10
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The Correct Option is B

Solution and Explanation

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
$H_{P}: H_{Q}: H_{R}: H_{S}$
$=\left(\frac{3}{4} i\right)^{2} \times 100:\left(\frac{3}{4} i\right)^{2} \times 10:\left(\frac{1}{4} i\right)^{2} \times 300$
$:\left(\frac{1}{4} i\right)^{2} \times 30$
$=30: 3: 10: 1$
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Concepts Used:

Kirchhoff's Laws

Kirchhoffs Circuit Laws allow us to solve complex circuit problems.

Kirchhoff's First Law/ Kirchhoffs Current Law

It states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. 

Kirchhoff's Second Law/ Kirchhoffs Voltage Law

It states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.