EE 2011 - Mid-Term Exam 1
October 10, 2012

Exam duration: 50 minutes

Instructions:
Closed-book, closed-notes.
Integration/graphing calculators allowed.
Detachable crib sheet and scratch page included.
Each problem is 10 points, there are four problems. Maximum score is 40 points.
Choose one of the four problems, and describe how you would solve it in words.
(No need to solve this chosen problem fully.)
Solve the remaining three problems fully.
Suggestion: Write down a line or two about how you would solve each problem at the very beginning.

Full Name:

Student ID Number:

Problem 1 (10 points)
Find the Th´venin equivalent circuit at terminals c-d for the circuit in Figure 1. e Figure 1:
10 Ω

k = 0.5

c

10 2cos(100t + 45o )
(in Volts)

100 mH

100 mH d 1

Working space for Problem 1

2

Problem 2 (10 points)

Figure 2:

1Ω

sin( 10 4 t )

0.1 mH

ZL

(in Amperes)

Load

(a) For the circuit in Figure 2, what should ZL be to ensure that maximum average power is delivered to the load (that is, average power dissipated by ZL is maximum)? Specify what components ZL would consist of, egs., a resistor and capacitor in series, or resistor and inductor, and what values these are.
(b) For the value of ZL you found in (a), what is the average power dissipated by the load ZL ?

3

Working space for Problem 2

4

Problem 3 (10 points)
If the current i(t) in Figure 3 has a phase angle of 10o , what is the value of the resistor R?
Figure 3:

100 mH cos(100 t + 45o )

i1 (t)

(in Volts)

5

R

Working space for Problem 3

6

Problem 4 (10 points)
We would like to adjust the load inductor L so that the voltage of the source v(t), and the current i(t), as indicated in Figure 4, have the same phase at ω = 104 rad/sec. What value of L will achieve this? Figure 4:

0.04 µF 1:10

v(t)

25 Ω

5:1

1Ω

L

i(t)

Load
Note: The load inductor L is NOT the same as the transformer coil impedance.

7

Working space for Problem 4

8

Scratch space

9

Phasors: v(t) = A cos(ωt + φo ) ↔ V = A∠φo
1
Impedance: R ↔ R, L ↔ jωL, C ↔ jωC
Complex Numbers: a + jb ↔ r∠φ or rejφ , where
√
b r = a2 + b2 , φ = tan−1 a and a = r cos φ, b = r sin φ a 1 b 1
2
a+jb = a2 +b2 − j a2 +b2 , j = −1, j = −j o o

o

o

1∠90o = ej90 = j, 1∠−90o = e−j90 = −j, 1∠0o = ej0 = 1, 1∠180o = ej180 = −1
Average Power: P = 0.5Vmax Imax cos(θv − θi ) watts
Power Factor (PF): cos(θv − θi ), if PF = 1, the load is purely resistive
Reactive Power: Q = 0.5Vmax Imax sin(θv − θi ) var
Reactive Factor (RF): sin(θv − θi ), if RF = 1, purely inductive, if RF = -1, purely capacitive
Complex Power: S = P + jQ, PF = cos (∠S), RF = sin (∠S)
1
1
RMS Voltage/Current: Vrms = √2 Vmax , Irms = √2 Imax
Source Transformation:
Voltage source V with impedance Z in series ↔ Current source I = (V /Z) with impedance Z in parallel
Maximum Power Transfer: If a source with impedance ZS drives a load of impedance ZL
∗
maximum power is transfered when ZL = ZS
If ZL = RL + jXL can only take restricted values, then choose XL to be as close to −XS
2
RS + (XL + XS )2 as possible

as possible, then choose RL as close to

If ∠ZL cannot be changed, but magnitude |ZL | can be changed, then choose |ZL | = |ZS |
Z22
Linear Transformer: Iprimary = Z11 Z22 +ω2 M 2 Vsource