Chapter 7, Problem 45E

(a)

To determine

Write the nodal equation of the circuit.

Answer to Problem 45E

The nodal equation of the circuit at node voltage $v_1$ is,

$C_1\frac{d{v}_1}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}-C_1\frac{d{v}_s}{dt}=0$ and at $v_2$ is,

$i_s+2\frac{v_2}{R}-\frac{v_1}{R}-\left(\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}\right)=0$

Explanation of Solution

Calculation:

The redrawn circuit is shown in Figure 1.

Here,

$v_s$ is the voltage supply across branch $AE$,

$C_1$ is the capacitor across branch $AB$,

$C_2$ is the capacitor across branch $BF$,

$R$ is the resistance across branch $BC$,

$R$ is the resistance across branch $CG$,

$i_s$ is the current supply from branch $DH$ and

$L$ is the inductor across branch $IJ$.

The expression for the current across the inductor is,

$i_L=\frac{1}{L}{\displaystyle {\int }_{t_0}^t{v}_{3}dt\text{'}}+i_L\left(t_0\right)$        (1)

Here,

$i_L$ is the current across inductor,

$v_3$ is the node voltage across inductor and

$L$ is the inductor in circuit.

The expression for the current across capacitor is,

$i_{C_1}=C_1\frac{d{v}_3}{dt}$        (2)

Here,

$i_{C_1}$ is the current across the capacitor and

$C_1$ is the capacitor in circuit.

The expression for the nodal analysis at node $v_1$ is,

$C_1\frac{d\left(v_1-v_3\right)}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}=0$        (3)

Rearrange equation (3),

$C_1\frac{d\left(v_1-v_3\right)}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}=0$

$C_1\frac{d{v}_1}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}-C_1\frac{d{v}_3}{dt}=0$        (4)

The voltage at node voltage $v_3$ is same as voltage source because of same node,

$v_3=v_s$        (5)

Substitute $v_s$ for $v_3$ in equation (4),

$C_1\frac{d{v}_1}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}-C_1\frac{d{v}_s}{dt}=0$

The expression for the nodal analysis at node voltage $v_2$ is,

$i_s+i_3=i_L+i_2$        (6)

The node voltage across resistance $R$ is $v_2$ because of same node.

$i_s+\frac{v_2}{R}=\frac{v_1-v_2}{R}+\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}$        (7)

Rearrange equation (7),

$\begin{array}{l}{i}_s+\frac{v_2}{R}-\left(\frac{v_1-v_2}{R}\right)-\left(\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}\right)=0\\ i_s+\frac{v_2}{R}+\frac{v_2}{R}-\frac{v_1}{R}-\left(\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}\right)=0\\ i_s+2\frac{v_2}{R}-\frac{v_1}{R}-\left(\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}\right)=0\end{array}$

Conclusion:

Thus, the nodal equation of the circuit at node voltage $v_1$ is $C_1\frac{d{v}_1}{dt}+C_2\frac{d{v}_1}{dt}+\frac{v_1-v_2}{R}-C_1\frac{d{v}_s}{dt}=0$ and at $v_2$ is

$i_s+2\frac{v_2}{R}-\frac{v_1}{R}-\left(\frac{1}{L}{\displaystyle {\int }_{t_0}^t\left(v_s-v_2\right)dt+i_L\left(t_0\right)}\right)=0$.

(b)

To determine

Write the mesh equation of the circuit.

Answer to Problem 45E

The mesh equation of the circuit in loop $\text{ABFEA}$ is $v_s-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t{i}_{1}dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$, in loop $\text{AIGDCBA}$ is $v_s-L\frac{d{i}_L}{dt}-\left(i_L-i_2\right)R-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$ and in loop $\text{BCGFB}$ is $2i_{2}R+\frac{1}{C_2}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt+v_s\left(t_0\right)}=0$.

Explanation of Solution

Calculation:

The redrawn circuit is shown in Figure 2 as follows,

Refer to the Figure 2,

The voltage at node voltage $v_3$ is same as voltage source because of same node $A$,

$v_3=v_s$        (8)

The expression for mesh equation in loop $ABFEA$ is,

$v_3-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t{i}_{1}dt}+v_s\left(t_0\right)-\frac{1}{C_2}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$        (9)

Here,

$v_3$ is the node voltage across branch $AE$,

$i_1$ is the current across the capacitor in branch $AB$,

$i_2$ is the current across the capacitor in branch $BF$,

$C_1$ and $C_2$ is the capacitor across branch $AB$ and $BF$ in circuit and

Substitute $v_s$ for $v_3$ in equation (9),

$v_s-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t{i}_{1}dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$

The expression for mesh equation in loop $AIGDCBA$ is,

$v_3-L\frac{d{i}_L}{dt}-\left(i_L-i_2\right)R-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$        (10)

Substitute $v_s$ for $v_3$ in the equation (10),

$v_s-L\frac{d{i}_L}{dt}-\left(i_L-i_2\right)R-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$

The expression for mesh equation in loop $BCGFB$ is,

$\begin{array}{l}-i_{2}R-i_{2}R-\frac{1}{C_2}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt+v_s\left(t_0\right)}=0\\ 2i_{2}R+\frac{1}{C_2}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt+v_s\left(t_0\right)}=0\end{array}$

Conclusion:

Thus, the mesh equation of the circuit in loop $\text{ABFEA}$ is $v_s-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t{i}_{1}dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$, in loop $\text{AIGDCBA}$ is $v_s-L\frac{d{i}_L}{dt}-\left(i_L-i_2\right)R-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt}+v_s\left(t_0\right)-\frac{1}{C_1}{\displaystyle {\int }_{t_0}^t\left(i_1-i_2\right)dt}+v_s\left(t_0\right)=0$ and in loop $\text{BCGFB}$ is $2i_{2}R+\frac{1}{C_2}{\displaystyle {\int }_{t_0}^t\left(i_2-i_1\right)dt+v_s\left(t_0\right)}=0$.