Question 212 of 380

Figure shows a part of an electric circuit. The potentials at points $a,b$ and $c$ are $30V,12V$ and $2V$ respectively. The current through the $20\Omega$ resistor will be,

A$1.0A$B$0.4A$C$0.6A$D$0.2A$

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Answer: (B) $0.4A$

Let the potential be $xV$ at the junction.
By using $V=IR$ we get,
$30-x=10{i}_{1} x-12=20{i}_{2} x-2=30{i}_{3}$
By using Kirchhoff's law,
${i}_{1}={i}_{2}+{i}_{3}$
$\Rightarrow \dfrac{30-x}{10}+\dfrac{12-x}{20}+\dfrac{2-x}{30}=0 \Rightarrow \dfrac{90-3x+18-1.5x+2-x}{30}=0$
$\Rightarrow x=20V$
The current through the $20\Omega$ resistor is
${i}_{2}=\left(\dfrac{30-12}{20}\right)A$
$\Rightarrow {i}_{2}=\dfrac{8}{20}A=0.4A$