Circuits

Exam 2 Solution

Fall 2018

1. /20

2. /25

3. /30

Extra Credit Exam 2

/10

4. /25

Extra Credit Exam 1

/10

Total /100

Name __________________

Please write your name at the top of every page!

Notes: 1) If you are stuck on one part of the problem, choose ‘reasonable’ values on the following parts to

receive partial credit.2) You don’t need to simplify all your numerical calculations. For example, you can leave square

root terms in radical form.

1

Name __________________

1) Short Answer (25 points)

1.1: Consider the Circuit A and Circuit B below (4 pts)

C2

R3R1

C1V1

R3R1

C1V1

R2

Circuit A Circuit B

T / F At DC steady state, Circuit A consumes no power

T / F At DC steady state, Circuit B consumes no power

T / F Circuit A can have a second order response

T / F Circuit B can have a second order response

F

T

F

T

1.2: If the voltage across a 2F capacitor is defined as determine the current through thecapacitor. (4 pts)

Vc t( ) 5 5e2 t=

Ic t( ) CdVC

dt= 20 e

2 t=

1.3: Which of the following transfer functions corresponds to a possible voltage measurement across an unknowncomponent in a second order RLC circuit with a step function source? (Circle all the correct answers) There are no'typos' in the expressions. (4 pts)

a) V s( )1

s s 4( ) s 8( )= c) V s( )

s

s 4( ) s 8( )=

a c

b) V s( )1

s 4( ) s 8( )= d) V s( )

1

s 4( ) s 8( )=

2

Name __________________

1.4: Which of the following transfer functions corresponds to a possible voltage measurement across an arbitrarycomponent for an underdamped second order RLC circuit with a step function source? (Circle all correct answers).There are no 'typos' in the expressions. (4 pts)

b ca) V s( )1

s 4( )2

= c) V s( )s

s2

2s 4=

b) V s( )1

s 4( )2

9= d) V s( )

s

s 2 4j( ) s 2 4j( )=

1.5: In the above figure, V1 is the voltage between node A and node B. IC is the current through the capacitor. Determinethe expression for V1(s) in terms of IC(s). All initial conditions are zero. Create one ratio but no need to simplify beyondthat. (4 pts)

R

C

L

1

2

A

B

V1

+

‐

V11

sC

R sLR sL

IC=

s2

RLC sL R

s2

LC sRCIC s( )

3

Name __________________

2) First Order Differential Equations (25 pts)

V19V

R1

6kohms

R26kohms

R36kohms

U1

TOPEN = 0

1 2

C12.5uF

I1

2mA

+

‐

vo(t)

2.1: Find vo(t) for t > 0. Note: The switch OPENS at t = 0. (15 pts)

Strategy: Find the Thevenin equivalent and use initial condtions to get final solution. Also some sourceconversion may be used to help reduce the circuit. Initial conditions depend on the switch opening.

V1 9V R1 6kΩ R2 6kΩ R3 6kΩ C1 2.5μF I1 2mA

When t>0 only I1, R3 and C1 are connected. Can convert source to VTH (currently looks like Norton)

VTH I1 R3 12V

RTH R3 6 103 Ω

τ RTH C1 0.015s

vo t( ) K1e

t0.015

K2=

To find K1 and K2 use initial and final conditions, final conditions first since it is DCSS

K2 12=

Need initial conditions which has the switch closed.

4

Name __________________

Make all parallel with source conversion then find thevenin

IV1

V1

R11.5 10

3 A

R1231

R1

1

R2

1

R3

12 10

3 Ω

Current looks like a norton circuit, convert source back to thevenin

VTH IV1 I1 R123 7V VC t0+( ) 7V=

7 K1 e0 12= K1 7 12 5

vo t( ) 12 5 e

t0.015= V

Vo(t)  [V] 

 

5

Name __________________

3) Second Order Differential Equations (30 pts)

V110V

R1

10 ohms

C10.01F

R210 ohms

V220V

L1

2H

1

2

U1

TCLOSE = 0

12

IL1(t)

3.1: Find IL1(t) for t>0 using Differential Equations

Find initial conditions. The circuit is at steady state due to 10V source. L1 is a short at steady state. C1 is open.

iL010V

10Ω= 1A= VC0 0V= shorted

When t>0:

Use thevenin then a source conversion to get it to a solved problem!

10V 20V( )10Ω

10Ω 10Ω 5 V This gives the voltage across R2 but you need to add 20 V to get Vth

VTH 20V 5V 15V

V115V

R1

5 ohms

C10.01F

L1

2H

1

2

RTH is two 10 ohm resistors in parallel so

RTH 5Ω

IL(t)  [A or mA]

 More space on the next page!

6

Name __________________

Really would prefer all components in parallel so source conversion!

R15 ohms

I13A

C10.01F

L1

2H

1

2

ωo1

2 0.017.071

α1

2 5 0.0110

α ω0 overdamped circuit

10 100 7.0712 2.929

10 100 7.0712 17.071

iL t( ) K1 e2.929 t K2 e

17.071 t K3=

t goes to infinity, steady state

K3 3=

first initial condition

iL 0( ) 1= K1 e2.929 t K2 e

17.071 t 3=

2 K1 K2=

second intial conditionC1

0

2

M1

2.929

1

17.071

1

diL 0( )

dt0= 2.929 K1 17.071K2=

M11

C12.414

0.414

iL t( ) 2.414 e

2.929 t 0.414 e17.071 t 3=

7

Name __________________

3.2: (Extra credit: 10 pts) Solve this above problem using Laplace transforms.

V110V

R1

10 ohms

C10.01F

R210 ohms

V220V

L1

2H

1

2

U1

TCLOSE = 01

2

IL1(t)

IL(t)  [A or mA] 

8

Graders: Check if students have the same answer, then walk through solving process to verify correct approach.

Name __________________

4) Second Order Laplace Transforms (25 pts)

+

‐Isw IL1

V19V

R1

250 ohms

R2

50 ohms

L11H

1

2

U1

TCLOSE = 0

12

C14uF

The switch in the figure above has been open for a long time and is closed at t = 0.

R1p3 250Ω R2p3 50Ω L1p3 1H V1p3 9V C1p3 4μF

4.1: Draw the s-domain equivalent circuit for t > 0. Include the initial conditions in your s-domain circuit. Please addvalues to your components. (5 pts)

for C11

s 4 106

2.5 105

1

s

IL 0-( ) IL 0+( )= 30mA=

for L1VC 0-( ) VC 0+( )= 0V= s 1 1s

S-domain equivalent circuit

R250 ohms

L1

1s

1

2

V1 = 0.03*1 = 0.03

C12.5E5 / s

V2 = 0

NOTE: Current direction is going upthrough the inductor (away from thevoltage source)....so the signs may bereversed if students assume traditionalconvention from top to bottom

9

Name __________________

4.2: Determine the current through the inductor as a function of time for t>0. Be sure to add your current direction toindicate your polarity. (20 pts)

C1R2

502.5 10

5s

502.5 10

5s

=1.25 10

7

s50 2.5 105

L1C1R21.25 10

7

s50 2.5 105

s= 1.25 107 s

250 2.5 10

5s

s 50 2.5 105

IL1

V1

L1C1R2=

0.03

1.25 107 s

250 2.5 10

5s

s 50 2.5 105

= Flip denominator.Divide top and bottom by 50

V1 = 0.03*1 = 0.03

L1C1R2

0.03 s 150

s2

5 103 s 2.5 10

5

s2

5 103 s 2.5 10

5 0=

4949.4897427831780982

50.510257216821901803

Partial fraction expansion

0.03 s 150s 4949( ) s 50.51( )

A1

s 4949( )

A2

s 50.51( )=

for A1:

0.03 4949( ) 1504949 50.51

3.123 104

IL(t)  [A or mA]

 

for A2:

0.03 50.51( ) 15050.51 4949

0.03

IL t( ) 30 e50.51 t

0.3123e4949 t= mA

10

Name __________________

Extra Credit (Exam 1) Do you want some extra credit for Exam 1? Complete the following problems that address bothUnit 1 and Unit 2 concepts. These will go toward your Exam 1 grade. NO PARTIAL CREDIT IS GIVEN!

U1

uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

L11 2

R

0VoutVin

EC 1 (5 pts): For the RL amplifier circuit, derive the relationship between Vout and Vin. As with RCamplifier circuits, KCL is a good starting point. You MUST include statements that define the ideal opamp to help solve for full credit. (The power is taken out for simplicity but the op amp is powered).

Vout =   

 

11

U1

uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

0V1 C1

1.59E-7

R1

2k

R2

1k

U2

uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

0R3

2k

L1

1.27H

1 2

V2

R4

3k

U3uA741

+3

-2

V+7

V-4

OUT6

OS11

OS25

R5

2k

R66k

0

Vout

EC 2 (5 pts): In the above circuit V1=V2=1sin(2π f t) where the frequency is 1 kHz. Determine Vout.

For clarity the following component values are restated below:

C1 1.59 107F R1 2kΩU1 op amp

R3 2kΩ L1 1.27HU2 op amp

U3 op amp R2 1kΩ R4 3kΩ R6 6kΩ R5 2kΩ

Vout =   

 

12