Download 2 sin 2 2 90 1 2.5 90 .4 2 90 2 90 2 90 1.5 164.3 1 3.32 15.7 3.2 1.6

Homework 9
1) Perform the calculations below. Put your answer in polar notation.
(2  j 3)(4  j )  3.6156.3(4.12  14.0)  14.8742.3
e j 30  e  j 24  0.866  j 0.5  0.914  j 0.41  1.78  j 0.09  1.782.89
1  2 j 2.24  63.4

 0.62  97.1
3 2 j
3.6133.7
j

j

j 5
e 2 (e 3 )  e 6  1150
322
 1.595
2  73
2) Use complex impedances to solve this circuit from Homework 8 for the steady state response,
vc. Do not solve for the transient response.
1
3.2
2 sin(2t)
2
0.8
0.2
Vc
Ans: You can approach the circuit as a voltage divider. The one branch, Z2 is the impedance
formed by the capacitor and the other branch, Z1, is the impedance formed by the series
combination of the resistor and inductor. Converting the input to its polar form and putting it
into the voltage divider equation gives:
2sin 2t  2  90
1


j


 Z2 
 2.5  90 
.4
vc  2  90 
  2  90 
  2  90 
  1.5  164.3
1
Z

Z
3.32


15.7


2
 1
 3.2  j1.6  j 
.4 

Or vc(t)= 1.5 cos(2t -164.3)
3) In the circuit below what are the currents in each of the elements in polar form? Graph them on
a single graph to show their relationship. What is the total current delivered by the source?
1
2H
1
2
cos2t
2
Ans: Since the elements are each across the voltage source with no other impedances in series,
we can quickly write the current for each element:
vs  10
10
 0.50  0.5  j 0
20
10
10
ic 

 290  0  j 2
1 0.5  90
j
2
10 10
il 

 0.25  90  0  j 0.25
j4
490
ir 
Summing the currents algebraically gives:
iTOTAL=0.5+j1.75
Graphing them gives:
ic
ir
il
4) In the circuit below, what is vo? Is the phase leading or lagging the input?
+
1
1
Vo
3
4cos4t
3
-
Ans: A good approach here would be to Thevenize the voltage source and the 1 and 3 ohm
resistors. This gives:
3
vTH  4    3
4
1(3)
RTH 
 0.75
(1  3)
The Thevenin resistance in series with the impedance of the capacitor forms one leg of a voltage
divider, Z1. The other 3 ohm resistor forms the second leg of the divider, Z2. We can then write
the output from this:
Z1  0.75  j
1
4
Z2  3  j0
 Z2 
30


vo  30 
  30 

 3.76  3.8 
 Z1  Z 2 
vo  2.43.8
5) In this circuit, find i. Then use that to find vL and vR. Show that these two voltages add up as
vectors to form the input, thus showing that KVL still holds.
1
2H
2cos2t
2
I
4
Ans: We can combine the impedance of the inductor in series with the resistor to form the total
impedance in the circuit. The current is then the voltage source divided by this impedance.
We can then find the voltage across the inductor and the resistor in turn by multiplying the
impedance of each element by the current (Ohm’s Law for impedances).
ZTOTAL  4  j 0  0  j 4  4  j 4  5.6645
20
 0.354  45
5.6645
vL  490(0.354  45)  1.41445
i
vR  40(0.354  45)  1.414  45
To show that the voltages add up as vectors:
vL
vR
vS
6) Use complex impedances to find vo.
+
1
2
1H
1/4
Vo
2
2
-
cos2t
Ans: The best way to tackle this would be to use nodal analysis. Writing the nodal equation
leads directly to a solution for vo:
1  vo
vo
v

 o
1
2
2  j2
j
0.5
1  vo  jvo  0.5vo  j 0.5v0
vo 
1
1

 0.632  18.4
1.5  j 0.5 1.5818.4