UNIVERSITY OF UTAH
ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT
HOMEWORK #6 Solution
ECE 1270
1.
Summer 2009
After being closed a long time, the switch opens at t = 0. Find i1 (t) for t > 0.
t=0
10kΩ
i1
10kΩ
60µA
3mH
Step 1: (Redraw circuit at t=0- and solve for iL. Inductor acts as a wire since it has sat for a long time)
10kΩ
60µA
i1
10kΩ
This circuit is a current divider:
60 µ ⋅10k
iL =
= 30 µ A
10k + 10k
iL
Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Inductor acts as a current
source since the current in the inductor has to remain the same.)
10kΩ
Only one current in the branch:
60µA
10kΩ
30µA
i1
i1 = −30 µ
Step 3: Final Value(Redraw circuit at t=∞
∞ and solve for unknown variable. Inductor acts as a wire
since it has sat for a long time in this position.)
10kΩ
There are no sources connected to the final
circuit:
i
i1
10kΩ
i1=0A
L
60µA
10kΩ
60µA
10kΩ
i1
To find Req the inductor is removed from the
final circuit to find path from top of inductor
to bottom of inductor:
L
3m
τ=
=
= 150n sec
R eq 10k + 10k
Step 4: Plug values into general equation:
i1 (t ) = 0 + [ −30µ − 0] e −t /150 n sec A = −30µ e−t /150 n sec A
1
2.
After being open for a long time, the switch closes at t = 0. Find V1(t) for t > 0.
t=0
2kΩ
20V +
+
V1
4kΩ
2µF
+
VC
Step 1: (Redraw circuit at t=0- and solve for VC. Capacitor acts as an open since it has been a long time)
2kΩ
There is no source connected between Vc so
V1 =VC=0
+
+
+
V1
20V
VC
4kΩ
Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Capacitor acts as a
voltage source since the voltage across capacitor has to remain the same.)
2kΩ
20V +
4kΩ
+
V1
0V
+
+ VC
V1 = 0V
∞ and solve for unknown variable. Capacitor acts as an open
Step 3: Final Value(Redraw circuit at t=∞
since it has sat for a long time in this position.)
2kΩ
V1 is found by a voltage divider:
20 ⋅ 4k 80k 40
+
+
V1 =
=
= V
2 k + 4k 6 k
3
V1
20V +
V
4kΩ
C
To find Req the capacitor is removed from the
final circuit(same circuit) to find path from top to
bottom of capacitor. Independent sources are removed and the equivalent resistance is found:
 1 
16m
τ = R eq ⋅ C = ( 4k 2k ) ⋅ 2µ =  1 1  ⋅ 2µ =
sec
6
 4k + 2k 
Step 4: Plug values into general equation:
40  40  −6t /16 m sec
40
V1 (t ) =
+ 0 −  e
V=
1 − e−6t /16 m sec V
3 
3
3
(
2
)
3.
After being open for a long time, the switch closes at t = 0. Find i1 (t) for t > 0.
2ki1
+ 8V
4mA
4kΩ
2µH
1kΩ
i1
4kΩ
Step 1: (Redraw circuit at t=0- and solve for iL. Inductor acts as a wire since it has sat for a long time)
iL
2ki1
+ 8V
When the circuit contains a dependent
source, an extra step is needed to determine
the value of the dependent variable:
+
4kΩ
4mA
1kΩ
-
8
= 1mA
8k
+
Taking a current summation at the top node
can be used to find iL gives:
iL = 2k ⋅ i1 + 4m ⇒ iL = 2k ⋅1m + 4m = 2.004 A
i1 -
i1
4kΩ
Note: −i1 ⋅ 4k + 8 − i1 ⋅ 4k = 0 ⇒ i1 =
Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Inductor acts as a current
source since the current in the inductor has to remain the same.)
V1
Solve the circuit for i1. Mesh currents or
node-voltage can be used. Node-voltage
2ki1
method:
+ 8V
4kΩ
2.004A
4mA
• With dependent sources: solve for
i1
1kΩ
4kΩ
dependent variable in terms of either
the mesh current or the node-voltage.
V 
i1 =  1 
 4k 
V −8 
−2k ⋅ i1 − 4m + 2.004 +  1
 + i1 = 0
 4k 
V 
 V − 8   V1 
−2k ⋅  1  − 4m + 2.004 +  1
+  = 0
 4k 
 4 k   4k 
1 
8
 −2 k 1
V1 
+
+  = +4m − 2.004 +
⇒
4k
 4 k 4k 4 k 
The desired variable is i1:
 4k 
V1 = −1.998 ⋅ 
 ≈ 4V
 −2.002k 
V  4
i1 =  1  =
= 1mA
 4k  4k
3
Step 3: Final Value(Redraw circuit at t=∞
∞ and solve for unknown variable. Inductor acts as a wire
since it has sat for a long time in this position.)
This wire puts 0V across 4k ohm resistor so:
i1=0
2ki1
+ 8V
4kΩ
4mA
1kΩ
i1
4kΩ
V1
2ki1
+
Vtest
4mA
-
1kΩ
+ 8V
4kΩ
Itest=
1A
i1
4kΩ
To find Req the inductor is removed from the
final circuit to find path from top of inductor
to bottom of inductor(Thevenin Resistance):
V
Using a test source: Rth = test
I test
Setting Itest=1A means that only Vtest
needs to be found.
Vtest = V1
V 
i1 =  1 
 4k 
V −8 
−2k ⋅ i1 − 4m − 1 +  1
 + i1 = 0
 4k 
V 
 V − 8   V1 
−2k ⋅  1  − 4m − 1 +  1
+  = 0
 4k 
 4k   4k 
1 
8
 −2k 1
V1 
+
+  = +4m + 1 +
4k
 4 k 4k 4 k 
Vtest = −2V
Rth =
 4k 
V1 = 1.006 ⋅ 
 ≈ −2V
 −2.002k 
⇒
Vtest −2V
=
= 2Ω
I test
1A
L
2µ
=
= 1µ sec
R eq
2
Step 4: Plug values into general equation:
i1 (t ) = 0 + [1m − 0] e−t /1µ sec A = 1me−t /1µ sec A
τ=
4
4.
After being open for a long time, the switch closes at t = 0. Find i1 (t) for t > 0.
2mF
5kΩ
ib
99i b
t=0
3kΩ
10V +
i1
10kΩ
2kΩ
Step 1: (Redraw circuit at t=0- and solve for VC. Capacitor acts as an open since it has been a long time)
+
VC
10V +
Solving for the dependent variable:
ib=0 which opens the dependent
source => 99ib=0
Taking a V-loop to get Vc value(Be CarefulIt is not 0 when there is a path with a V src.)
+0 + 10 + VC − 0 = 0 ⇒ VC = −10V
5kΩ
ib
99i b
3kΩ
10kΩ
i1
2kΩ
Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Capacitor acts as a
voltage source since the voltage across capacitor has to remain the same.)
Using node-voltage to solve this circuit to find i1:
First find dependent variable in terms of nodevoltage variable, V1.
ib
V1 − ( −10) − ( −( −10)) V1
+
5kΩ
i
=
=
+
b
-10V
VC
10k
10k
99i b
Next, take current summation equation at V1 node:
3kΩ
V1 V1 (V1 − V2 )
10V +
+ −
=0
V2
10k 3k
2k
Current summation equation at V2 node:
i1
10kΩ
2kΩ
(V2 − V1 ) − 99 ⋅ V1 − ( 0 − V2 ) = 0
2k
10k
5k
V1
Using the first equation and solving for V1:
Plugging into the second equation:
V1 V1 V1 V2
V2  −15V2   1 
1  −15V2  V2
+ −
=
−
− 99 ⋅
=0




+
10k 3k 2k 2k
2 k  4   2k 
10k  4  5k
20
30  V2
 6
 1 15 99(15) 1 
V1 
+
−
V2  + +
+  = 0 ⇒ V2 = 0
=
60
k
60
k
60
k
2
k


40k
5k 
 2 k 8k
V1 =
V2  60k  −15V2

=
2k  −4 
2
V1 = 0
i1 = 0
5
Step 3: Final Value(Redraw circuit at t=∞
∞ and solve for unknown variable. Capacitor acts as an open
since it has sat for a long time in this position.)
ib=0
+
VC
99i b
Since i1 is always the same => i1(t)=0.
i1
10kΩ
5.
i1=0
3kΩ
10V +
this means that 99ib source becomes open.
5kΩ
ib
2kΩ
Using superposition, derive an expression for V that contains no circuit quantities other than
is , R1, R 2 , R 3 , R 4 , α, or Vs .
αi x
is
ix
+ Vs
R1
R4
+
V
R2
R3
Step 1: is=0(off), Vs=on
αi x
V1
i2 = ix
+ Vs
R1
i1
+
V
Using mesh currents:
i1 = −α ix
+
R2
ix
i2
+
R3
+
R4
Using a voltage loop and then substituting above eq:
( +i1 − i2 ) R3 + VS − i2 R4 = 0
( −α ix − ix ) R3 + VS − ix R4 = 0
ix ( +α R3 + R3 + R4 ) = VS
ix =
VS
( +α R3 + R3 + R4 )
V = α ix R2 =
Step 1: is=on, Vs=0(off)
αi x
αVS R2
( +α R3 + R3 + R4 )
Using mesh currents:
i1 = is
i2 = −α ix
+
is
i1
+
V
R1
R2
ix
i2
i3
+
i3 = ix
+
R4
R3
+
6
Using a voltage loop and then substituting above eq:
( +i2 − i3 ) R3 − i3 R4 = 0
( −α ix − ix ) R3 − ix R4 = 0
ix ( +α R3 + R3 + R4 ) = 0
ix = 0
i2 = 0
V = ( +i1 − i2 ) R2 = is R2
The total V is the sum of both solutions:
αVS R2
V = is R2 +
( +α R3 + R3 + R4 )
6. After being closed for a long time, the switch opens at t=0.
a) Calculate the energy stored on the inductor as t -> ∞.
b) Write a numerical expression for v(t) for t> 0.
20kΩ
t=0
300 mV
+
30kΩ
+
v(t)
150 mH
7
3kΩ
7. After being open for a long time, the switch closes at t=0.
a) Write an expression for vc(t=0+).
b) Write an expression for vc(t>0) in terms of
is , R1, R 2 , R 3 , and C.
t=0
R1
is
C
vC +
R3
8
R2
3kΩ
Use the circuit below for both problem 8 and 9.
8.
Calculate the value of RL that would absorb maximum
power.
+
3 mA
24kΩ
2kΩ
22kΩ
9.
Calculate that value of maximum power RL could absorb.
9
15V
RL
10.
Using superposition, derive an expression for i that contains no circuit quantities other than
is , R1, R 2 , R 3 , R 4 , α, or Vs .
is
R1
R2
i
αi x +
R4
R3
10
ix
vs +