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Practice Final Exam Solution: Problem 1 - Overdamped Response in RLC Circuits, Exams of Electrical and Electronics Engineering

Material Type: Exam; Class: Fund Electric Circuits; Subject: Electrical & Computer Engg; University: University of Utah; Term: Unknown 1989;

Typology: Exams

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2260
N. Cotter
PRACTICE FINAL EXAM SOLUTION: Prob 1
1. (50 points)
C
R
t = 0
A
+
V = 100 V
2
R
L
-
-
After having been open for a long time, the switch is closed at t = 0.
R1 = 12.5R2 = 12.5L = 6.25 µH
a. Two capacitances are available: 250 nF and 2 nF. Specify the value of C
that will make v(t) overdamped.
b. Using the value of C found in (a), write a time-domain expression for v(t).
ans: a) C = 250 nF
b) v(t) = 13.3 (e-0.4Mt – e–1.6Mt) V
sol'n: (a) To make the response overdamped, we must have two real characteristic
roots. We use the circuit for t > 0, consisting of C, R1, L, and vA in series.
We may find the characteristic equation by looking it up in a textbook or by
setting the impedance of R1, C, and L in series to zero.
z=R1+1
sC +sL =s2+R1
L
s+1
LC =0
The characteristic roots for the quadratic equation are
s1,2 =R
2L±R
2L
2
1
LC
or
s1,2 =−α ±α2 ωo
2α R
2Lωo1
LC
We want an overdamped response, (real roots α2 > ωo2).
α=R
2L=12.5Ω
(2) 6.25 µH=12.5
12.5 M rad/s =1 M rad/s
Try each C value in turn.
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Download Practice Final Exam Solution: Problem 1 - Overdamped Response in RLC Circuits and more Exams Electrical and Electronics Engineering in PDF only on Docsity!

N. Cotter PRACTICE FINAL EXAM SOLUTION: Prob 1

  1. (50 points)

C

R

t = 0

A

V = 100 V

R

L

v

After having been open for a long time, the switch is closed at t = 0.

R

1

= 12.5Ω R

2

= 12.5Ω L = 6.25 μH

a. Two capacitances are available: 250 nF and 2 nF. Specify the value of C

that will make v(t) overdamped.

b. Using the value of C found in (a), write a time-domain expression for v(t).

ans: a) C = 250 nF

b) v(t) = 13.3 (e

-0.4Mt

  • e

–1.6Mt ) V

sol'n: (a) To make the response overdamped, we must have two real characteristic

roots. We use the circuit for t > 0, consisting of C, R 1

, L, and v A

in series.

We may find the characteristic equation by looking it up in a textbook or by

setting the impedance of R 1

, C, and L in series to zero.

z = R

1

sC

+ sL = s

2

R

1

L

s +

LC

The characteristic roots for the quadratic equation are

s

1 ,

R

2 L

R

2 L

2

LC

or

s

1 ,

2

o

2

R

2 L

o

LC

We want an overdamped response, (real roots α

2

ω o

2 ).

R

2 L

(2) 6.25 μH

M rad/s = 1 M rad/s

Try each C value in turn.

C = 2 nF:

o

6.25 μH ⋅ 2 nF

12.5 m ⋅ 1 μ

1 M

rad/s

ω o

= 8.9M rad/s > α

2 (underdamped)

C = 250 nF:

o

6.25 μH ⋅ 250 nF

1562.5 m ⋅ 1 μ

1 M

rad/s

ω o

= 0.8M rad/s < α

2 (overdamped)

We need C = 250 nF for an overdamped solution.

sol'n: (b) We use the exponential solution for the overdamped case:

v ( t ) = A

1

e

s 1

t

+ A

2

e

s 2

t

+ A

3

Because the value of A 3

is all that is left of v(t) as t → ∞, we first find the

constant term, A 3

. (The other terms decay because the characteristic roots

always have negative real parts in a passive RLC circuit. When the switch

opens, the energy sloshing back and forth in the L and C will decay owing to

power dissipated by the series resistor R 1

As t → ∞, the circuit reaches equilibrium. C acts like an open circuit, L acts

like a short circuit or wire.

Model:

R

v C

i L

V

A

= 100 V

v C

(∞) = 100 V

i L

(∞) = 0 A

  • v(∞)–

Since L acts like a wire, there is no voltage drop across it.

Thus, A 3

= v(t→∞) = 0.

We find coefficients A 1

and A 2

by matching initial conditions in the circuit.

We find initial conditions by examining the circuit at t = 0

  • , when the circuit

has reached equilibrium. We find the values of i L

and v C

, the energy

variables, at t = 0

  • and use the same values at t = 0

(since the energy in the

circuit cannot change instantly).

v (

) = v

A

− i

L

) R

1

− v

C

) = 100 V − 4 A⋅12.5Ω − 50 V = 0 V

The same equation applies for t > 0, and we may differentiate to find dv(t)/dt

in terms of energy (or state) variables i L

and v C

v ( t ) = v

A

− i

L

( t ) R

1

− v

C

( t )

dv ( t )

dt

di

L

( t )

dt

R

1

dv

C

( t )

dt

The basic equations for L and C, rearranged, allow us to translate the

derivatives on the right side of this equation into non-derivatives we can

calculate numerically.

di

L

( t )

dt

L

v

L

( t )

dv

C

( t )

dt

C

i

C

( t )

Applying these identities, we have

dv ( t )

dt

L

v

L

( t ) R

1

C

i

C

( t ).

Only now that we have differentiated do we finally evaluate the derivative

we seek at t = 0:

dv ( t )

dt

t = 0

L

v

L

) R

1

C

i

C

dv ( t )

dt

t = 0

6.25μH

⋅ 0 V⋅12.5Ω −

250 nF

i

C

Since i C

is in series with i L

, we have i C

) = i L

) = 4A.

dv ( t )

dt

t = 0

4 A

250 nF

= − 16 MV/s

Now we find A 1

and A 2

v (

) = 0 = A

1

+ A

2

⇒ A

2

= − A

1

dv ( t )

dt

t = 0

= − 16 MV/s = A

1

s

1

+ A

2

s

2

= A

1

( s

1

− s

2

s

1

− s

2

2

o

2

2

o

2

2

o

2

= 2 ( 1 M )

2

− (0.8 M )

2

= ( 2 ) 0.6 M = 1.2 M

Concluding the algebra, we find the numerical values of the coefficients A 1

and A 2

A

1

16 M v/s

1.2 M

= 13.3 v/s

A

2

= −13.3 v/s

Using the values of α and ω o

from above, we find the values of s 1

and s 2

s 1

= –1 M + 0.6 M = –0.4 M

s 2

= –1 M – 0.6 M = –1.6 M

Plugging into the general form of underdamped solution completes our

answer:

v(t) = 13.3 (e

-0.4Mt

  • e

–1.6Mt ) V