Question 2.16: Applying KCL, determine current IS in the electric circuit t......

Now applying Kirchhoff’s current law to nodes A and B we have

I_1=I_2+I_s               (i)

and                I_2+I_3=\frac{V_1}{4}            (ii)

also voltage of node B = V_0 =16 V
Voltage across AC + voltage across AB = voltage at node B.

Solving eq (i), (ii), (iii), and (iv) we have

Therefore,                           I_{\mathrm{s}}=1 \mathrm{~A} \quad I_3=\frac{V_1}{4}-I_2=3-1=2 \mathrm{~A}