In the circuit shown in the figure, the value of node voltage 𝑉
In the circuit shown in the figure, the value of node voltage 𝑉2 is
A. 22 + j2 V
B. 2 + j 22 V
C. 22 – j 2 V
D. 2 – j 22 V
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Since there is a voltage source present between the two nodes, the concept of supernode will be applied.
By KCL, using supernode at Node (1) and (2), we can write:
\(4 = \frac{{{V_1}}}{{ - j3}} + \frac{{{V_2}}}{6} + \frac{{{V_2}}}{{j6}}\)
\(4 = \frac{{j2{V_1} + {V_2} - j{V_2}}}{6}\)
24 = j2 V1 + (1 - j) V2 ---(1)
By KVL, V1 – V2 = 10 ∠0°
V1 = V2 + 10 ---(2)
Substituting (2) in (1), we get:
24 = j2 (V2 + 10) + (1 - j) V2
24 = j2 V2 + j20 + (1 - j) V2
24 = j20 + (1 + j) V2
\({V_2} = \frac{{24 + j20}}{{1 + j}} = \frac{{24 - j20}}{{\left( {1 + j} \right)}} \times \frac{{\left( {1 - j} \right)}}{{\left( {1 - j} \right)}}\)
\({V_2} = \frac{{\left( {24 - j20} \right)}}{2} \times \left( {1 - j} \right)\)
\({V_2} = \frac{{24 - j20 - j24 - 20}}{2}\)
\({V_2} = \frac{{4 - j44}}{2}\)
V2 = 2 – j22 V
