## Q. 1.P.7

If R = 10 Ω, find $V_2$ in figure given below.

(a) 22 V (b) 23.4 V (c) 10 V (d) 21.4 V

## Verified Solution

Apply nodal analysis and KCL at node 1,
we have

$\frac{V_1-100}{20}+\frac{V_1}{20}+\frac{V_1-V_2}{10}=0$      (i)

Applying KCL at node 2, we have

$\frac{V_2}{30}+\frac{V_2}{30}+\frac{V_2-V_1}{10}=0$         (ii)

Solving Eq. (i), we have

\begin{aligned} & V_1-100+V_1+2 V_1+2 V_2=0 \\ & \Rightarrow 4 V_1-2 V_2=100 \end{aligned}

From Eq. (ii), we get

$\begin{gathered} V_2+V_2+3 V_2-3 V_1=0 \\ -3 V_1+5 V_2=0 \\ {\left[\begin{array}{rr} 4 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} V_1 \\ V_2 \end{array}\right]=\left[\begin{array}{r} 100 \\ 0 \end{array}\right]} \\ \Delta_1=\left[\begin{array}{rr} 100 & -2 \\ 0 & 5 \end{array}\right]=500, \Delta_2=\left[\begin{array}{rr} 4 & 100 \\ -3 & 0 \end{array}\right]=300 \text { and } \\ \Delta=\left[\begin{array}{rr} 4 & -2 \\ -3 & 5 \end{array}\right]=20-6=14 \end{gathered}$

We have

$V_1=\frac{\Delta_1}{\Delta}=\frac{500}{14}=35.7 V \text { and } V_2=\frac{\Delta_2}{\Delta}=\frac{300}{14}=21.4 V$