On this question, I am supposed to find Iphi and V1. I Applied KCL at Node 1, and made an auxillary equation for the supernode? Should I have left all the terms in fractions for my matrix-vector equation? Where did I go wrong? Example 10: Determine \( v_1 \) and \( i_{\phi} \) in the circuit shown in Figure 19.[Figure showing a circuit diagram]Figure 19: Circuit for Example 10.The circuit diagram consists of:- A 60 V voltage source.- Resistors labeled as 2 \( \Omega \) and 3 \( \Omega \) on the upper path.- A dependent current source labeled as \( 6i_{\phi} \).- Parallel paths with resistors labeled as 24 \( \Omega \) and another 3 \( \Omega \).- Nodes labeled \( V_1 \) and \( V_2 \), with \( v_1 \) indicated across the 24 \( \Omega \) resistor.**Auxiliary Equation:**\[ V_2 - 6i_{\phi} = 60 \]**Given:**We know what \( v_3 \) is, so we don’t need to solve for it.**KCL at \( V_1 \):**\[ \frac{V_1 - 60}{2} + \frac{V_1}{24} + \frac{V_1 - V_2}{3} = 0 \]- LCM is 24, leading to: \[ 12V_1 - 720 + V_1 + 8V_1 - 8V_2 = 0 \] \[ 21V_1 - 8V_2 = 720 \] *(Eq. 1)***Auxiliary Equation:**\[ V_2 - 6\left( \frac{V_2 - V_1}{2} \right) = V_3 \]\[ V_2 - \frac{6V_2 - 6V_1}{3} = 60 \]\[ 3V_2 - 6V_2 + 6V_1 = 180 \]\[ -6V_1 - 3V_2 = 180 \] *(Eq. 2)*Conclusion:- Solve Equations 1 and 2 to find \( v_1 \) and \( i_{\phi} \).- Note \( V_3 = 60 \). This image contains a mathematical calculation involving matrices and determinants commonly found in linear algebra, particularly in solving systems of linear equations using Cramer's rule.1. **Matrix Equation:** \[ \begin{bmatrix} 21 & -8 \\ -6 & -3 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} 720 \\ 180 \end{bmatrix} \] This represents a system of linear equations with unknowns \( V_1 \) and \( V_2 \).2. **Determinant Calculation:** The determinant \(\Delta\) of the coefficient matrix is computed as: \[ \Delta = 21 \cdot (-3) - (-6) \cdot (-8) = -111 \]3. **Cramer's Rule:** - **Determinant \(\Delta_1\):** Substitute the first column of the coefficient matrix with the constants: \[ \Delta_1 = \begin{vmatrix} 720 & -8 \\ 180 & -3 \end{vmatrix} = 720 \cdot (-3) - 180 \cdot (-8) = -720 \] - **Determinant \(\Delta_2\):** Substitute the second column of the coefficient matrix with the constants: \[ \Delta_2 = \begin{vmatrix} 21 & 720 \\ -6 & 180 \end{vmatrix} = 21 \cdot 180 - (-6) \cdot 720 = 8100 \]4. **Solving for Unknowns:** Using Cramer's rule: \[ V_1 = \frac{\Delta_1}{\Delta} = \frac{-720}{-111} = \frac{240}{37} \quad \text{(Note: should get 480)} \] \[ V_2 = \frac{\Delta_2}{\Delta} = \frac{8100}{-111} = -\frac{

The equivalent circuit with assumed node current direction is shown below: From the equivalent circuit, the current I∅ is given by: In the equivalent circuit, apply KVL between node 3 and ground node and apply KVL between node 2 and node 3: In the equivalent circuit, apply nodal analysis at node 1:The matrix equation for equation 4 and equation 3 is given by: From the matrix equation the values of voltage at node 1 and voltage at node 2 are given by: Thus, the current in 3Ω is given by: Hence, the voltage at across the 24Ω resistor is 48 V and the current in 3Ω resistor flowing from right to left is -4 A.