Consider the following circuit, with given constants ?-40V, R = 20?, L-4mH, C,-5
ID: 2304829 • Letter: C
Question
Consider the following circuit, with given constants ?-40V, R = 20?, L-4mH, C,-5uP and C,-5?? where the switch S has been closed for long enough so that the circuit has reached a steady state, at which point the switch is opened (a) Use Kirchoff's Law to obtain a differential equation for the circuit, expressed only in terms of the charge q in the circuit and other given constants, and write down the initial conditions (current and voltage across each element) at the time the switch is opened. (Hint: Try to simplify the circuit first.) (b) Using the general form of the solution to the differential equation you wrote down for part (a), for q(t), and the initial conditions, obtain an expression for the current through R at any given time t. (c) What are the natural frequency and the oscillation frequency of the circuit? (d) Obtain an expression for the energy (not power) dissipated as heat in resistor R as a function of time (e) What is the energy lost as heat during the first ten milliseconds (from t Oms to t-10ms) right after the switch is opened? What is the energy dissipated as heat during the second ten milliseconds (from t-10ms to t-20ms) after the switch is opened?Explanation / Answer
given
E = 40 V
R = 20 ohm
L = 4 mH
C1 = 5uF
C2 = 5uF
hence in parallel connection
Ceff = C1 + C2 = 10 uF
at steadt state, at t = 0, switch is opened
a. at t = 0
the circuit is in steady state
hence, L acts like short circuit, C acts like open circuit
hence
charge on each capacitor = qo = Q
Q = C1*E = C2*E
also
E = IoR ( where Io is current through resistor)
when the switch is opened
from kirchoff's voltage law
let charge on each capacitor at time t be q
then
-q/C1 + Ldi/dt + iR = 0
where i = 2dq/dt
hence
q/C1 = Ldi/dt + iR
q/C1 = 2Lq" + 2Rq'
qo = Q = C1E = C2E
b. hence
the solution to the above ODe is
q(t) = c1*e^-2500(1 + sqrt(5))t + c2*e^2500(sqrt(5) - 1)t
at t = 0
5*10^-6*40 = c1(1) + c2(1) = c1 + c2
q'(t) = -2500(1 + sqrt(5))*c1*e^-2500(sqrt(5) + 1)t + 2500(sqrt(5) - 1)c2*e^2500(sqrt(5) - 1)t
at t= 0, i = 0
hence
-2500(1 + sqrt(5))*c1 + 2500(sqrt(5) - 1)c2 = 0
sc2(sqrt(5) - 1) = c1(sqrt(5) + 1)
c2 = c1*(sqrt(5) + 1)^2/(5 - 1)
c2 = c1(3 + sqrt(5))/2
c1 + c2 = c1 ( 5 + sqrt(5))/2 = 5*10^-6*40
c1 = 5.52786*10^(-5)
c2 = 1.447213*106(-4)
hence
q'(t) = -0.447213268*e^(-2500(sqrt(5) + 1)t) + 0.447213411*e^(2500(sqrt(5) - 1)t) = i(t)/2
i(t) = -0.894426536*e^(-2500(sqrt(5) + 1)t) + 0.89442682*e^(2500(sqrt(5) - 1)t)
c. natural frequency, w = 1/sqrt(LCeff) = 5000 rad/s
f = w/2*pi = 795.774715 Hz
d. energy dissipated in resistor R as a funciton of time is
E(t) = i(t)^2*R*t
E(t) = [(-0.894426536*e^(-2500(sqrt(5) + 1)t) + 0.89442682*e^(2500(sqrt(5) - 1)t))]^2*R*t
E(t) = 20[-0.894426536*e^(-2500(sqrt(5) + 1)t) + 0.89442682*e^(2500(sqrt(5) - 1)t)]^2*t
e. for the forst 10 ms
energy lost = integrate(i^2(t)*R*dt)
E = integrate( 20[(-0.894426536*e^(-2500(sqrt(5) + 1)t) + 0.89442682*e^(2500(sqrt(5) - 1)t))]^2 dt)
E = -0.000988853*e^(-16180.3t) + 0.00639999*e^(-5000t) + 0.0025885*e^(6180.34t) + constant
hence
E = (-0.000988853*e^(-16180.3*0.01) + 0.00639999*e^(-5000*0.01) + 0.0025885*e^(6180.34*0.01)) - (-0.000988853 + 0.00639999 + 0.0025885)
E = 1.79466071*10^24 joules
similiarly
for the next 10 ms
E = 1.244*10^51 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.