The value of current through the 20,Omega res | vedclass

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Question

(A) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at this node:$\frac{V-10}{10} + \frac{V-(-6)}{20} + \frac{V

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(A) Let the potential at the central node be $V$. Applying Kirchhoff's Current Law $(KCL)$ at this node:$\frac{V-10}{10} + \frac{V-(-6)}{20} + \frac{V}{10} + \frac{V-10}{5} = 0$Multiplying the entire equation by $20$:$2(V-10) + (V+6) + 2V + 4(V-10) = 0$$2V - 20 + V + 6 + 2V + 4V - 40 = 0$$9V - 54 = 0$$V = 6 \,V$Now,the current through the $20\,\Omega$ resistor is given by:$I = \frac{V - (-6)}{20} = \frac{6 + 6}{20} = \frac{12}{20} = 0.6 \,A$.

The value of current through the $20\,\Omega$ resistor is ............ $A$.

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