Answer to Question #240527 in Differential Equations for mds

Question

In an R-L-C series circuit, the differential equation for the instantaneous charge q(t) on the capacitor is   2 2   d q dq q L R Et dt dt C . Determine the charge q(t) and current i(t) for a circuit with R  10 ohm, L = 1 henry, C = 2 10 farad and E(t) = 50 10 cos t volts. What is the steady-state current for this circuit?

Answer 1

$$1\dfrac{d^2q}{dt^2}+10\dfrac{dq}{dt}+\dfrac{q}{10^{-2}}=50\cos(10t)$$

Homogeneous differential equation

$$\dfrac{d^2q}{dt^2}+10\dfrac{dq}{dt}+100q=0$$

Corresponding (auxiliary) equation

$$r^2+10r+100=0$$

$$D=(10)^2-4(1)(100)=-300$$

$$r=\dfrac{-10\pm\sqrt{-300}}{2(1)}=-5\pm 5\sqrt{3}i$$

The general solution of the homogeneous differential equation is

$$q_h=c_1e^{-5t}\cos(5\sqrt{3}t)+c_2e^{-5t}\sin(5\sqrt{3}t)$$

Find the particular solution of the non homogeneous differential equation

$$q_p=A\cos(10t)+B\sin(10t)$$

$$q_p'=-10A\sin(10t)+10B\cos(10t)$$

$$q_p''=-100A\cos(10t)-100B\sin(10t)$$

Substitute

$$-100A\cos(10t)-100B\sin(10t)-100A\sin(10t)$$

$$+100B\cos(10t)+100A\cos(10t)+100B\sin(10t)$$

$$=50\cos(10t)$$

$$100B=50$$

$$A=0$$

The particular solution of the non homogeneous differential equation

$$q_p=\dfrac{1}{2}\sin(10t)$$

The general solution of the given differential equation

$$q(t)=c_1e^{-5t}\cos(5\sqrt{3}t)+c_2e^{-5t}\sin(5\sqrt{3}t)+\dfrac{1}{2}\sin(10t)$$

Then

$$i(t)=\dfrac{dq}{dt}=-5c_1e^{-5t}\cos(5\sqrt{3}t)-5c_2e^{-5t}\sin(5\sqrt{3}t)$$

$$-5\sqrt{3}c_2e^{-5t}\sin(5\sqrt{3}t)+5\sqrt{3}c_2e^{-5t}\cos(5\sqrt{3}t)$$

$$+5\cos(10t)$$

$t\to\infin$

 The steady-state current for this circuit is

$$i_{steady-state}=5\cos(10t)$$