Q.
When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be :
Solution:
Let voltage at $C = xv$
KCL : $i_1 + i_2 = i$
$\frac{20-x}{2} + \frac{10-x}{4} = \frac{x-0}{2}$
$ \Rightarrow x = 10 $
and $i = 5\, amp$.
