Example 12
R = 200, L =5, C = .001, E(t) = 100sin(10t), E’(t) = 1000cos(10t), Isp = Acos(ωt)+Bsin(ωt). Find Isp in the form Isp(t) = I0sin(ωt-δ).
Voltage loop equation
5I^''+200I^'+1000I=1000cos(10t)
Steady periodic current
I_sp=A cos〖(10t)+B sin(10t) 〗
〖I'〗_sp=-10A sin(10t)+10B cos(10t)
〖I''〗_sp=-100A cos〖(10t)-100B sin(10t) 〗
Plugging Isp and its derivatives into the voltage loop equation and simplifying yields
500A cos(10t)+2000B cos(10t)-2000A sin(10t)+500B sin(10t)=1000cos(10t)
500A+2000B=1000
-2000A+500B=0
A=2/17 B=8/17
I_sp=2/17 cos〖(10t)+8/17 sin(10t) 〗=2/17 [cos〖(10t)+4 sin(10t) 〗 ]
δ=〖2π-tan〗^(-1)(1/4)=6.038
Magnitude of Isp
I_o=√(1^2+4^2 )=√17
I_sp (t)=2/√17 sin(10t-6.038)
Alternatively, we could find the magnitude Io and phase angle using the formulas
I_o=E_o/Z=E_o/√(R^2+(ωL-1/ωC)^2 )
δ=tan^(-1)((LCω^2-1)/ωRC)
The above equations are derived from the geometry of right triangles like the method previously used. It can also be noted that the current magnitude resembles Ohm’s Law I = V/R. The value Z is known as the impedance which is a way to calculate the resistance to current flow for devices that don’t have a simple resistance value like a resistor. The resistance provided by inductors and capacitors is shown in the above equation to be a function of the angular frequency of the power source. At very low and high frequencies, relative to the values of L and C you choose, inductors and capacitors respond very differently than they do around the resonance frequency due to the nonlinear nature of the impedance. That is, they have an operational bandwidth in which they can output without a reduced gain. This property is the reason that capacitors and inductors see wide usage in filter circuits. By choosing the correct values for the components you can significantly reduce the magnitude of signals at frequencies you don’t want expressed in your output.
P.S. I am still slightly boozed up and have been up for almost 24 hours.
