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Chapter 3

Q. 3.2

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

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

Step-by-Step

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 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 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 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 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.

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