250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0
250 Ω w i250 + V₁(t) i 500 500 Ω www x₁ = V₂ (+) and x₂ = 12 (4) V₁ (+) - 1250 (+) · 25012 - Vc (+) =0 (+) sin (2000+) a) KVL: 1,50 (+) = KCL: V₁(+)-vc(+) 250♫2 isto (t) + i₂ (+) + C.dv₂ (+) C.dv (+) = 0 1µF 0.5 H& → 1500 (+) = VC (+) and c = 1 µF = 1×10-6F 50052 X, <+> dx₁(+) H C == ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+)) 5000 The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250 resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and the source turns on at time t = 0. (a) Find the state equation for this circuit. (b) Find the output equation. (c) Using your state and output equations, use MATLAB to plot the resistor currents for 0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth. = = 1 1x10-bp (Xx(t) - X₁ (+) 5000 (1x10 bp) (X₂ (+)- X₁ (+)) (-2000)x, (+) + (106) x₂ (+) - X₂ (+): dx₂ (t) = L· div(t) dt = = = Vect) -isto (+). 500 V₁ (+) - X₁ (+) - 500 X₂ (+) 0.5 H Yo.5 (V. (+) == x₁ (+)-(500)xx (+)) 2 (sin (2000) - X₁(+) -500x₂ (+)) 2 sin (2000t) -2x, (+) -(1000)x, (+)) A 54 Matux: x=Ax+Bu B dx, (+) -2000 106 X₁(+) 0 = dr (f) Sin (2000) -2 1800 Xx(+) b) Output Vastables: Y₁ = 1250 (+) - Y₁ = 1500 (+) = V₁ (+) - √c (+) 2502 Sin (2000) - x, (+) VC(+) 500 = X₁ (+) 50022 output Mataxi 3 ८ + D 4(+) = (x+Du (५.) 250 0 X₁ (+) = Sin(2000) Y₂(+) 0 500 Xx (+) 0 750-0
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter19: Capacitors
Section: Chapter Questions
Problem 1PP: Fill in all the missing values. Refer to the formulas that follow. Resistance Capacitance Time...
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I worked on this problem for a long time but apparently the analysis was not right. Can you show what the correct way is to solve? (Only Part A and B, not C.)
Transcribed Image Text:250 Ω
w
i250
+
V₁(t)
i 500
500 Ω
www
x₁ = V₂ (+) and x₂ = 12 (4)
V₁ (+) - 1250 (+) · 25012 - Vc (+) =0
(+) sin (2000+)
a)
KVL:
1,50 (+) =
KCL:
V₁(+)-vc(+)
250♫2
isto (t) + i₂ (+) + C.dv₂ (+)
C.dv (+)
= 0
1µF
0.5 H&
→ 1500 (+) = VC (+)
and
c = 1 µF = 1×10-6F
50052
X, <+>
dx₁(+)
H
C
== ₁₁ (+) = 1500 (+) = ↓ (iN(+) - X₁ (+))
5000
The source voltage is v₁(t) = sin(2000t). We wish to obtain the current through the 250
resistor (1250) and the 500 2 resistor (1500). There is no initial energy stored in the circuit and
the source turns on at time t = 0.
(a) Find the state equation for this circuit.
(b) Find the output equation.
(c) Using your state and output equations, use MATLAB to plot the resistor currents for
0≤t ≤ 4 ms. Use a sufficient number of samples so that the plot appears smooth.
=
=
1
1x10-bp
(Xx(t) - X₁ (+)
5000
(1x10 bp) (X₂ (+)- X₁ (+))
(-2000)x, (+) + (106) x₂ (+)
- X₂ (+): dx₂ (t) = L· div(t)
dt
=
=
= Vect) -isto (+). 500
V₁ (+) - X₁ (+) - 500 X₂ (+)
0.5 H
Yo.5 (V. (+) == x₁ (+)-(500)xx (+))
2 (sin (2000) - X₁(+) -500x₂ (+))
2 sin (2000t) -2x, (+) -(1000)x, (+))
A
54 Matux:
x=Ax+Bu
B
dx, (+)
-2000 106
X₁(+)
0
=
dr (f)
Sin (2000)
-2
1800
Xx(+)
b)
Output Vastables:
Y₁ = 1250 (+) -
Y₁ = 1500 (+) =
V₁ (+) - √c (+)
2502
Sin (2000) - x, (+)
VC(+)
500
=
X₁ (+)
50022
output Mataxi
3
८
+
D
4(+) = (x+Du
(५.)
250
0
X₁ (+)
=
Sin(2000)
Y₂(+)
0
500
Xx (+)
0
750-0
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