Preliminary This laboratory will demonstrate how varying resistance will change
ID: 1806588 • Letter: P
Question
Preliminary
This laboratory will demonstrate how varying resistance will change the natural response of a series RLC circuit (Figure 1). The signal generator is modeled as an ideal voltage source where vs (t) = 5 u(t) V in series with an source resistance RS = 50 ohm. The inductor will be modeled as an ideal L = 80 mH inductor in series with a wire resistance Rwire = 30 ohm. An LCR meter was used to determine these values. The capacitance is C = 0.1 microF.
Calculate roots of the characteristic equation, and determine the general form of the natural
response of the capacitor voltage vC,n (t) for t ? 0 (i.e., overdamped, critically damped, and
underdamped) when the resistor Rvar has a value of 4700 ohm.
Find vC (0), d vC /d t|t = 0, and vC (?) given no initial energy is stored in the inductor and/or capacitor.
Explanation / Answer
KVL yields :
Vs = (Rs + Rwire + Rvar)i + Vc + Ldi/dt
let Rs + Rwire + Rvar = R
since i = C dVc/dt s you get :
LC d2Vc/dt2 + RC dVc/dt + Vc = Vs
dividing by LC gives :
d2Vc/dt2 + (R/L) dVc/dt + (1/LC)Vc = Vs/LC
roots :
=> s2 + (R/L)s + (1/LC) = 0
with the given values of R,L,and C you get :
s2 + 59750s + 125x106 = 0
=> discriminant = 59750^2 - 4(125x106) > 0
so it's overdamped
=> s1,2 = [-59750 ± (discriminant)] / 2
=> s1 = -2170.9 , s2 = - 57579
so natural response is :
Vcn(t) = K1 e-2170.9t + K2 e- 57579t
Forced response is :
Vcf(t) = A = Vs = 5
=> Vc(t) = 5 + K1 e-2170.9t + K2 e- 57579t.
Vc(0) = 0 = 5 + K1 + K2
=> K1 + K2 = -5 ...... (1)
and Vc'(0) = 0 = -2170.9 K1 - 57579K2 = 0
=> K1 = -26.52 K2 ... (2)
hence from (1) :
-26.52 K2 + K2 = - 5
=> K2 = 0.196
=> K1 = -5.196
so Vc(t) = 5 - 5.196 e-2170.9t + 0.196 e- 57579t.