The figure shows a part of an electric circui | vedclass

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(A) Let the potential at the junction point be $x$.Applying Kirchhoff's Current Law $(KCL)$ at the junction point,the sum of currents leaving the junc

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$0.4$

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(A) Let the potential at the junction point be $x$.Applying Kirchhoff's Current Law $(KCL)$ at the junction point,the sum of currents leaving the junction must be zero:$\frac{x-30}{10} + \frac{x-12}{20} + \frac{x-2}{30} = 0$Multiplying the entire equation by $60$ to clear the denominators:$6(x-30) + 3(x-12) + 2(x-2) = 0$$6x - 180 + 3x - 36 + 2x - 4 = 0$$11x - 220 = 0$$11x = 220$$x = 20\,V$Now,the current through the $20\,\Omega$ resistor is given by:$I = \frac{x - 12}{20}$$I = \frac{20 - 12}{20} = \frac{8}{20} = 0.4\,A$.

The 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 $........\,A$.

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