Download ECE1250F14_PracticeEx1p2soln

1250
F 14
EX:
PRACTICE EXAM 1 PROBLEM 2 SOLUTION
a) Find a symbolic expression for v3 in the circuit below using not more than
symbolic component names is, R1, R2,  = 32 (in the dependent voltage source), and
 = 4 (in the dependent current source).
b) Find the numerical value of v3 in the circuit below.
SOL'N: a) Because the problem says nothing about what method of solution must be
used, we might use any of the tools studied thus far: Ohm's law, KVL,
KCL, voltage-divider, current-divider, Thevenin source transformation, or
Norton source transformation.
The latter four methods require special configurations, which are lacking
in this circuit. Although is and R2 are a Norton form, converting to a
Thevenin form would cause ix to disappear, making the circuit unsolvable.
A similar argument applies to the Thevenin form consisting of 32is and
R1. Current iy would disappear if a Norton form were used.
So we revert to using Ohm's law, KVL, and KVL, which always works.
We label voltage drops and currents for all resistors.
Starting with v-loops, we skip the left side inner loop, owing to the
presence of current source is. The other inside loops are skipped for the
same reason. The only larger loop that lacks a current source is in the
second and fourth branches from the left:
Using the sign of the voltage measurement where we enter a component,
we have the following KVL equation:
-v1 - a ix + v2 = 0V
The circuit has only two essential nodes, shown in red and green below:
(1)
We look first for components in series carrying the same current. This
happens in the second branch with R1 and the dependent voltage source,
but we are unconcerned about the current in the voltage source, so we
move on.
We get equivalent equations from both essential nodes. If we sum the
currents out of the top node, we get the following current summation:
-is + iy + biy + ix = 0 A
(2)
Now we apply Ohm's law, using labeled currents already provided.
v1
R1
(3)
v
v2 = ix R2 or ix = 2
R2
(4)
v1 = iy R1 or iy =
Since we are solving for a voltage, we opt for replacing resistor current
terms with resistor voltage terms.
Equation (1) becomes
v
-v1 - a 2 + v2 = 0V
R2
(5)
Equation (2) becomes
v
v
v
-is + 1 + b 1 + 2 = 0 A
R1
R1 R2
(6)
Where is v3? The answer is that, since v3 is a voltage drop for a current
source, we will first solve the circuit. Then we will find v3 using a voltage
loop. Note that it is fine to include a voltage loop including v3 at any time.
The only issue is that it increases the number of equations we must deal
with.
Our next step is to put (5) and (6) in standard form, meaning we put
variables on the left and constants on the right:
æ
ö
1
(-1)v1 + ç -a
+1÷ v2 = 0V
R2 ø
è
(7)
æ 1
æ 1ö
1ö
çè R + b R ÷ø v1 + çè R ÷ø v2 = is
1
1
2
(8)
and
Looking ahead, v3 is in parallel with v2, so solving for v2 is what we want
to do. Solving (7) for v1, we have the following equation:
æ
ö
1
- ç -a
+ 1÷ v2
R2 ø
æ
ö
è
1
v1 =
= ç -a
+ 1÷ v2
(-1)
R2 ø
è
(9)
Now we substitute for v1 in (8).
æ 1
ö
æ 1ö
1 öæ
1
çè R + b R ÷ø çè -a R +1÷ø v2 + çè R ÷ø v2 = is
1
1
2
2
(10)
Factor out v2 on the left side.
éæ 1
ö æ 1 öù
1 öæ
1
+1÷ + ç ÷ ú v2 = is
êç + b ÷ ç -a
R1 ø è
R2 ø è R2 ø û
ëè R1
(11)
or
v2 =
is
æ 1
ö æ 1 ö
1 öæ
1
çè R + b R ÷ø çè -a R + 1÷ø + çè R ÷ø
1
1
2
2
(12)
Voltage v3 is in parallel with v2, so v3 = v2.
v3 = v2 =
is
æ 1
ö æ 1 ö
1 öæ
1
çè R + b R ÷ø çè -a R +1÷ø + çè R ÷ø
1
1
2
2
(12)
We get some simplification by multiplying top and bottom by R1R2.
v3 =
is
RR
× 1 2
æ 1
ö æ 1 ö R1R2
1 öæ
1
çè R + b R ÷ø çè -a R + 1÷ø + çè R ÷ø
1
1
2
2
(12)
or
v3 =
is R1R2
(1+ b ) ( -a + R2 ) + R1
b) We plug in the numbers.
v3 =
1.5 A(100 W)(12 W)
1.8 kV
1.8 kV
=
=
= ¥V
0
(1+ 4 ) ( -32 +12 W ) +100 W 5(-20) +100
Interesting. This circuit cannot be realized with the values given!
(12)