The series RLC circuit shown in Fig. 7.18 has the following parameters: C = 0.04 F, L = 1 H, R = 6 Ω, iL(0)=4 A, and vC(0)=-4 V. The equation for the current in the circuit is given by the expression

Question

The series RLC circuit shown in Fig. 7.18 has the following parameters: C = 0.04 F, L = 1 H, R = 6 Ω, [latex]i_L[/latex](0)=4 A, and [latex]v_C[/latex](0)=-4 V. The equation for the current in the circuit is given by the expression

[latex]\frac{d^{2} i}{d t^{2}}+\frac{R}{L} \frac{d i}{d t}+\frac{i}{L C}=0[/latex]

A comparison of this equation with Eqs. (7.14) and (7.15) illustrates that for a series RLC circuit the damping term is R/2L and the undamped natural frequency is [latex]1 / \sqrt{L C}[/latex]. Substituting the circuit element values into the preceding equation yields

[latex]\frac{d^{2} x(t)}{d t^{2}}+2 \zeta \omega_{0} \frac{d x(t)}{d t}+\omega_{0}^{2} x(t)=0[/latex] 7.14

[latex]s^{2}+2 \zeta \omega_{0} s+\omega_{0}^{2}=0[/latex] 7.15

[latex]\frac{d^{2} i}{d t^{2}}+6 \frac{d i}{d t}+25 i=0[/latex]

Let us determine the expression for both the current and the capacitor voltage.

Accepted Answer

The characteristic equation is then

[latex]s_{2}+6 s+25=0[/latex]

and the roots are

[latex]s_{1}=-3+j 4 [/latex]

[latex]s_{2}=-3-j 4[/latex]

Since the roots are complex, the circuit is underdamped, and the expression for i(t) is

[latex]i(t)=K_{1} e^{-3 t} \cos 4 t+K_{2} e^{-3 t} \sin 4 t[/latex]

Using the initial conditions, we find that

[latex]i(0)=4=K_{1}[/latex]

and

[latex]\frac{d i}{d t}=-4 K_{1} e^{-3 t} \sin 4 t-3 K_{1} e^{-3 t} \cos 4 t+4 K_{2} e^{-3 t} \cos 4 t-3 K_{2} e^{-3 t} \sin 4 t[/latex]

and thus

[latex]\frac{d i(0)}{d t}=-3 K_{1}+4 K_{2}[/latex]

Although we do not know di(0)/dt, we can find it via KVL. From the circuit we note that

[latex]\operatorname{Ri}(0)+L \frac{d i(0)}{d t}+v_{C}(0)=0[/latex]

or

[latex]\frac{d i(0)}{d t} =-\frac{R}{L} i(0)-\frac{v_{C}(0)}{L} [/latex]

[latex]=-\frac{6}{1}(4)+\frac{4}{1} [/latex]

=-20

Therefore,

[latex]-3 K_{1}+4 K_{2}=-20[/latex]

and since [latex]K_{1}=4, K_{2}=-2[/latex], the expression then for i(t) is

[latex]i(t)=4 e^{-3 t} \cos 4 t-2 e^{-3 t} \sin 4 t \mathrm{~A}[/latex]

Note that this expression satisfies the initial condition i(0)=4. The voltage across the capacitor could be determined via KVL using this current:

[latex]R i(t)+L \frac{d i(t)}{d t}+v_{C}(t)=0[/latex]

or

[latex]v_{C}(t)=-R i(t)-L \frac{d i(t)}{d t}[/latex]

Substituting the preceding expression for i(t) into this equation yields

[latex]v_{C}(t)=-4 e^{-3 t} \cos 4 t+22 e^{-3 t} \sin 4 t \mathrm{~V}[/latex]

Note that this expression satisfies the initial condition [latex]v_{C}(0)=-4 \mathrm{~V}[/latex].

The MATLAB program for plotting this function in the time interval t > 0 is as follows:

Question 7.8: The series RLC circuit shown in Fig. 7.18 has the following ...

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