In the Wheatstone's network given, \(P = 10\,\Omega , Q = 20\,\Omega , R = 15 \,\Omega , S = 30 \,\Omega\), the current passing through the battery (of negligible internal resistance) is
The correct answer is A:\(0.36A\) The balanced condition for Wheatstone's bridge is \(\frac{P}{Q}=\frac{R}{S}\) as is obvious from the given values. No, current flows through galvanometer is zero. Now, \(P\) and \(R\) are in series, so Resistance \(R_{1}=P+R\) \(=10+15=25 \,\Omega\) Similarly, \(Q\) and \(S\) are in series, so Resistance \(R_{2}=R+S\) \(=20+30=50 \,\Omega\) Net resistance of the network as \(R_{1}\) and \(R_{2}\) are in parallel \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) \(\therefore R=\frac{25 \times 50}{25+50}=\frac{50}{3} \,\Omega\) Hence, \(I=\frac{V}{R}=\frac{6}{50 / 3}=0.36\, A\)
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.
