Figure shows a part of an electric circuit. The potentials at points $a , b$ and $c$ are $30\,V , 12\,V$ and $2\,V$ respectively. The current through the $20 \Omega$ resistor will be $........\,A$
JEE MAIN 2023, Diffcult
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sum of current at junction point will be zero:
$\frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0$
$\Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30}$
$\Rightarrow x\left(\frac{6+3+2}{60}\right)=\frac{180+36+4}{60}$
$\Rightarrow x =\frac{220}{11}=20\,V$
$\therefore \text { Current through } 20\,\Omega=\frac{x-12}{20}$
$=\frac{20-12}{20}=\frac{2}{5}=0.4\,A$
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