Question 6.6: Circuit Modeling Using the Loop Method: Two Loops Reconsider...
Circuit Modeling Using the Loop Method: Two Loops
Reconsider the circuit in Figure 6.21 and solve Example 6.4 using the loop method.
Assign loop currents as shown in Figure 6.23. Note that there are two loops with unknown currents.
For loop 1, applying Kirchhoff’s voltage law gives
v_{\mathrm{L}}+v_{\mathrm{R}}-v_{\mathrm{a}}=0.
Expressing the voltage across each element in terms of the loop currents, we have
L \frac{\mathrm{d} i_{1}}{\mathrm{~d} t}+R\left(i_{1}-i_{2}\right)-v_{\mathrm{a}}=0.
Similarly, applying Kirchhoff’s voltage law to loop 2 gives
\begin{gathered} v_{\mathrm{C}}+v_{\mathrm{R}}=0, \\ \frac{1}{C} \int i_{2} \mathrm{~d} t+R\left(i_{2}-i_{1}\right)=0 . \end{gathered}
Differentiating the above equation with respect to time results in
\frac{1}{C} i_{2}+R \frac{\mathrm{d} i_{2}}{\mathrm{~d} t}-R \frac{\mathrm{d} i_{1}}{\mathrm{~d} t}=0.
The differential equation obtained in each loop is first-order; thus, the circuit is a second-order system. The two first-order differential equations can be written in matrix form as follows:
\left[\begin{array}{cc} L & 0 \\ -R & R \end{array}\right]\left\{\begin{array}{c} \mathrm{d} i_{1} / \mathrm{d} t \\ \mathrm{~d} i_{2} / \mathrm{d} t \end{array}\right\}+\left[\begin{array}{cc} R & -R \\ 0 & \frac{1}{C} \end{array}\right]\left\{\begin{array}{c} i_{1} \\ i_{2} \end{array}\right\}=\left\{\begin{array}{c} v_{\mathrm{a}} \\ 0 \end{array}\right\} .
Note that the output voltage v_{\mathrm{o}} is related to the current through the capacitor, that is, v_{\mathrm{o}}=\frac{1}{C} \int i_{2} \mathrm{~d} t or V_{\mathrm{o}}(s)=\frac{1}{C s} I_{2}(s). Taking the Laplace transform of the system of differential equations gives
\left[\begin{array}{cc} L s+R & -R \\ -R s & R s+\frac{1}{C} \end{array}\right]\left\{\begin{array}{l} I_{1}(s) \\ I_{2}(s) \end{array}\right\}=\left\{\begin{array}{c} V_{\mathrm{a}}(s) \\ 0 \end{array}\right\}.
Using Cramer’s rule to solve for I_{2}(s), we have
I_{2}(s)=\frac{R s}{R L s^{2}+\frac{L}{C} s+\frac{R}{C}} V_{\mathrm{a}}(s).
Thus,
V_{\mathrm{o}}(s)=\frac{1}{C s} I_{2}(s)=\frac{R}{R L C s^{2}+L s+R} V_{\mathrm{a}}(s).
Taking the inverse Laplace transform gives the same input-output equation obtained previously in Example 6.4,
R L C \ddot{v}_{\mathrm{o}}+L \dot{v}_{\mathrm{o}}+R v_{\mathrm{o}}=R v_{\mathrm{a}}.
