EE 223
Test #1
02/25/2003
Name: ____________________________________
_____ / 100 pts
Open book - Closed notes - Show all of your steps in detail to get full credit!
1. Example:
(30 pts)
For the circuit shown, find:
iR(t)
3u(-t) mA
vC(t)
+
–
R=6kΩ
1 mA
a) Value of the time constant τ
b) Is this a natural or step
response problem?
c) vc(t) for t ≥ 0
d) iR(t) for t ≥ 0
1/12*10-3F
a) τ = R C = 6kΩ 1/12 mF = 0.5 s
b) Step response due to changing current source 3u(-t) mA
c) and d):
iR(0−) = 3 mA + 1 mA = 4 mA
vC(0−) = vC(0+) = vR(0) = iR(0−) R = 4 mA 6 kΩ = 24 V
vC(∞) = vR(∞) = iR(∞) R = 1 mA 6 kΩ = 6 V
vCN(t) = A e-t/τ
vCF = 6 V ⇒
vC(t) = vCF + vCN(t)
⇒
vC(t) = 6V + (24V − 6V) e-t/0.5s = 6V + 18V e-2t for t ≥ 0
⇒
iR(t) = vC(t) / R = 6V/6kΩ + 18V/6kΩ e-t/0.5s = 1mA + 3mA e-2t for t ≥ 0
25
15
C
voltage V (V)
20
10
5
0
-1
0
1
2
3
4
5
6
time (s)
τ = __500 ms_
_____step response problem ________
vc(t) = ___6V + 18V e-2t_____________ iR(t) = ____1mA + 3mA e-2t____________
2. Example
(40 pts)
0.8H
iL(t)
250Ω
The switch in the circuit shown
has been open for a long time
before closed at t = 0. Find:
+
250Ω
–
t=0
+
−
vC(t)
a) What kind of damping is
present in this circuit?
b) iL(0−) and iL(0+)
c) vC(0−) and vC(0+)
d) vC as t → ∞
e) vC(t) for t ≥ 0
f) iL(t) for t ≥ 0
C=5µF
6V
3
a) Parallel RLC - circuit:
α = 1 / (2RC) = 1 / (2 250Ω 5µF) = 400 s-1
2
1.5
C
voltage V (V)
ω0 = 1 / (√LC) = 1 / (√ 0.8H 5µF) = 500 s-1
2.5
α < ω0 ⇒ underdamped
ωd = √(ω02 - α2) = 300 s-1
b) iL(0−) = iL(0+) = −6V / (2*250Ω) = −12 mA
c) vC(0−) = vC(0+) = 6V / 2 = 3 V
1
0.5
0
-0.5
0
2
4
6
8
10
tim e (m s )
12
14
16
18
20
d) no source in RLC-circuit ⇒ vC(∞) = 0 V, only natural response fN(t) ≠ 0 (also: iL(∞)=0A)
e) vC(t) = e-αt [ A1 cos(ωdt) + A2 sin(ωdt) ] + 0V
t = 0+: A) vc(0+) = 3 V = A1
B) dvc(0+)/dt = ic(0+)/C = (-iL - iR)/C = (12mA - 3V/250Ω) = 0A and
dvc(0+)/dt = -αe-αt [ A1 cos(ωdt) + A2 sin(ωdt) ] +e-αt [-ωdA1sin(ωdt)+ωdA2cos(ωdt)]
= -αA1 + ωdA2
0 = -αA1 + ωdA2 ⇒ A2 = αA1 / ωd = 400 3V/300 = 4V
vC(t) = e-400t [ 3 cos(300t) + 4 sin(300t) ] V for t ≥ 0
f) iL(t) = e-αt [ A3 cos(ωdt) + A4 sin(ωdt) ]
t = 0+: A) iL(0+) = -12 mA = A3
B) diL(0+)/dt = vL(0+)/L = 3V / 0.8H = 3.75 A/s = -αA3 + ωdA4 ⇒ A4 = -3.5 mA
iL(t) = e-400t [ -12 cos(300t) - 3.5 sin(300t) ] mA for t ≥ 0
____________underdamped_________
vC(∞) = ____0 V____
iL(0−) = __-12 mA__ iL(0+) = __-12 mA_ vC(0−) = __3 V____ vC(0+) = ____3 V___
vc(t) = _e-400t [ 3 cos(300t) + 4 sin(300t) ] V_ iL(t) = _e-400t [ -12 cos(300t) - 3.5 sin(300t) ] mA_
3. Example
(30 pts)
For the circuit shown, find:
a) The AVERAGE power
delivered by the source (in Watts)
+
~
−
50Ω
10µF
0.5H
500 √2 cos(377t) V
b) The REACTIVE power
delivered by the source (in VAR)
c) The APPARENT power
delivered by the source (in VA)
d) Find the Power Factor and tell
whether it is leading or lagging.
(How do you know?)
ω = 2πf = 377 s-1
Z = R || XL || XC = 50 || jω0.5H || -j / ω10µF = 50 || j188.5Ω || -j265.25Ω =
= 49.71 + j3.82Ω = 49.85∠4.39° Ω
ϕ = 4.39°
IEFF = VEFF / Z = 500V / 49∠4.39°Ω = 10.03∠-4.39°A = 10 - j0.768 A
P = VEFF IEFF cos(ϕ) = 5 kW
Q = VEFF IEFF sin(ϕ) = 384 VAR
S = VEFF IEFF = 5015 VA
PF = cos(ϕ) = 0.977 lagging (current lags voltage)
P = __5 kW__ Q = ___384 VAR___ S = __ 5015 VA__ PF = __0.977 lagging__