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## Q. 3.2

Find the voltage drop in each resistor of the circuit in Fig. 3.15.

## Verified Solution

Each resistor is treated as an equivalent spring element; accordingly, the circuit in Fig. 3.15a is decomposed in the equivalent spring system in Fig. 3.15b. It

Table 3.4 Similarity of a spring system with other engineering systems

 Engineering problems Equivalent terms in a spring element Stiffness Load Field variable U k F $\frac{EA}{L}$ P u $\frac{GJ}{L}$ T θ $\frac{\Pi D^{4} }{128\mu }$ $\varrho$ P $\frac{1}{R}$ I V $\frac{KA}{L}$ q T

consists of 4 elements and 4 nodes. Table 3.5 shows the result of decompositions and corresponding element models.
Assembling the element models in Table 3.5 gets the system model as

$\begin{bmatrix} 0.2 & -0.2 & 0 & 0 \\ -0.2 0.2 & +0.1 +0.05 & -0.1 -0.05 & 0\\ 0 & -0.1-0.05& 0.1+0.05+0.1 & -0.1 \\ 0 & 0 & -0.1 & 0.1 \end{bmatrix}\begin{bmatrix} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{bmatrix}=\begin{bmatrix} I_{1.1} \\ I_{2.1} + I_{2.2} + I_{2.3} \\ I_{3.2} + I_{3.3} + I_{3.4} \\ +I_{4.4} \end{bmatrix}\leftarrow \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix}$  (3.12)

Table 3.5 The decomposition of equivalent spring system

 Element Node i Node j ke (1/R) Element model 1 ➊ ➋ 0.2 $\begin{bmatrix} 0.2 & -0.2 \\ -0.2 & 0.2\end{bmatrix} \begin{bmatrix} V_{1} \\ V_{2} \end{bmatrix} = \begin{bmatrix} I_{1.1} \\ I_{2.1} \end{bmatrix}$ 2 ➋ ➌ 0.1 $\begin{bmatrix} 0.1 & -0.1 \\ -0.1 & 0.1\end{bmatrix} \begin{bmatrix} V_{2} \\ V_{3} \end{bmatrix} = \begin{bmatrix} I_{2.2} \\ I_{3.2} \end{bmatrix}$ 3 ➋ ➌ 0.05 $\begin{bmatrix} 0.05 & -0.05 \\ -0.05 & 0.05\end{bmatrix} \begin{bmatrix} V_{2} \\ V_{3} \end{bmatrix} = \begin{bmatrix} I_{2.3} \\ I_{3.3} \end{bmatrix}$ 4 ➌ ➍ 0.1 $\begin{bmatrix} 0.1 & -0.1 \\ -0.1 & 0.1\end{bmatrix} \begin{bmatrix} V_{3} \\ V_{4} \end{bmatrix} = \begin{bmatrix} I_{3.4} \\ I_{4.4} \end{bmatrix}$

The circuit system has two boundary conditions: (1) the current at nodes 1 and 4 are given as 1 A and − 1 A, respectively, and no additional current is input at node 2 and node 3; (2) node 4 is grounded as a reference $(V_{4}= 0_{v})$. Therefore, the system model Eq. (3.12) can be simplified as

$\begin{bmatrix} 0.2 & -0.2 & 0 & 0 \\ -0.2 & 0.35 & -0.15 & 0 \\ 0 & -0.15 & 0.25 & -0.1 \\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$  (3.13)

Solving Eq. (3.13) obtains the solution at [V] = $[21.6667, 16.6667, 10, 0]^{T}$, and the voltage drops over the resistors $R_{1}, R_{2}, R_{3}$, and $R_{4}$ are 5, 6.6667, 6.6667, and 10 voltage, respectively.