EE 201 ELECTRIC CIRCUITS
Class Notes
CLASS 3
The material covered in this class will be as follows:
⇒
Kirchoff’s Voltage Law.
⇒
Fundamental Laws of Electric Circuits.
⇒
Dependent Voltage Source.
⇒
Dependent Current Source.
At the end of this class you should be able to:
⇒
Apply Kirchoff’s Voltage Law.
⇒
Recognize invalid circuits.
⇒
Use the fundamental laws to analyze electric circuits.
⇒
Recognize the symbol of a dependent source.
⇒
Distinguish between the four possible types of dependent sources.
⇒
Analyze circuits that contain dependent sources.
Kirchoff’s Voltage Law (KVL):
The algebraic sum of voltages around any closed circuit is equal to zero.
KVL around circuit 1 (CW)
⇒
−v1 − v2 + v3 − v4 + v5 = 0 (1)
KVL around circuit 1 (CCW)
⇒
+ v1 + v2 − v3 + v4 − v5 = 0
(2) [same as (1)]
CW = clockwise & CCW = counterclockwise
KVL around the outer circuit (CW)
KVL around circuit 2 (CW)
⇒
⇒
−v6 + v8 + v3 − v4 + v5 = 0
−v6 + v7 = 0
⇒
(3)
v6 = v7 (parallel elements)
Figure 1
Alternative KVL Statement:
The algebraic sum of voltages between two nodes is independent of the path taken from the first node
to the second node.
path1&2
KVL
Node a
KVL
Node a
→
Node b
⇒
Node b
⇒
path 2&3
→
+ v2 + v1 = + v3 − v4 + v5
+v3 − v4 + v5 = −v8 + v6
(4)
[same as (1)]
(5)
[same as (3)]
Figure 2
Example 1:
Calculate the unknown voltages in the given circuit.
Ω
Ω
Ω
Figure 3
Solution:
Applying KVL:
Right-hand circuit (CW)
⇒
−(7) + v1 + (−1) + 10 = 0
⇒
v1 = −2V
Right-hand circuit (CCW)
⇒
+ (7) − (10) − (−1) − v1 = 0
⇒
v1 = −2V
⇒
+ v1 = + (7) − (10) − (−1)
⇒
v1 = −2V
path1&2
Node a
→
Node b
Same answer in all cases.
Left-hand circuit (CW)
path 3&4
Node a
→
Node c
⇒
+ (7) − (v2 ) = 0
⇒
v2 = 7V
⇒
+ v2 = +7
⇒
v2 = 7V
Same answer in both cases.
Ω
Ω
Ω
Ω
Ω
Ω
Figure 4
Fundamental Laws of Electric Circuits:
1- Ohm’s Law, KCL and KVL are the fundamental laws of electric circuits.
2- All the fundamental laws of electric circuits must be satisfied.
3- If a given circuit violates at least one of the fundamental laws, the circuit is not valid.
Example 2:
All the given circuits below are invalid. Why?
Ω
Ω
Ω
Ω
Ω
Figure 5
Solution:
⇒
4+2 = 7
a)
KCL at node a ⇒
b)
KVL around left hand circuit ⇒ +10 + 6 + 4 = 0 ⇒ 20 = 0 ⇒
c)
KCL at node a ⇒
2=0
⇒
KCL not satisfied
KVL not satisfied
KCL not satisfied
KVL around lower part of the circuit (CW) ⇒ 12 = 0 ⇒
Example 3: In the given circuit, calculate the unknown quantities.
Ω
Ω
Ω
Figure 6
KVL not satisfied
Solution:
-
V2 +
7Ω
I1
-
a
+
V1
-
10V
V4 +
6Ω
2A
4Ω
-
3A
Vs
V3 +
Figure 7
KCL at node a ⇒ 3 = 2 + I1 ⇒
Ohm’s law ⇒
I1 = 1A
v2 = +7 I1 = 7 × 1 = 7V
KVL around left hand circuit ⇒ v1 + 10 − v2 = 0 ⇒ v1 + 10 − 7 = 0 ⇒ v1 = −3V
Ohm’s law ⇒
v3 = −3 × 4 = −12V
+ v4 + v1 − v3 − vs = 0
(KVL around right hand circuit)
⇓
+ (3 × 6) + v1 − v3 − vs = 0
(Ohm’s law)
⇓
+18 + (−3) − (−12) − vs = 0
⇒
vs = 27V
Ideal Dependent Sources.
A voltage source whose voltage depends on another voltage or current is called a dependent voltage
source.
Symbol
Figure 8
A current source whose current depends on another voltage or current is called a dependent current
source.
Symbol
Figure 9
Example 4:
Circuit (a) I s depends on v2
⇒
I s (voltage-dependent current source)
Circuit (b) vs depends on v1
⇒
vs (voltage-dependent voltage source)
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 10
Four possible types of dependent sources:
Voltage-dependent voltage source (it is a voltage source that depends on another voltage)
Current-dependent voltage source (it is a voltage source that depends on another current)
Voltage-dependent current source (it is a current source that depends on another voltage)
Current-dependent current source (it is a current source that depends on another current)
Example 5:
a) Calculate the value of the dependent current source.
b) Show that the power generated is equal to the power dissipated.
Ω
Ω
Figure 11
Solution:
a) KCL at node a
⇒
−i − i2 + 4i = 0
−v2 + v1 + 20 = 0
⇒
i2 = 3i
(KVL around left hand circuit)
⇓
−(40i2 ) + (20i ) + 20 = 0 (Ohm’s law)
⇓
−(40 × 3i ) + 20i + 20 = 0 [Using (1)]
⇓
−120i + 20i + 20 = 0
∴ 4i = 4 ×
⇒
i=
20 1
= A
100 5
1 4
= A (Value of the dependent current source)
5 5
1 20
b) p20V = +20i = +20 × =
= 4W
5 5
1
1
1
4
p20 Ω = +iv1 = + ( )(20 × ) = + ( )(4) = W
5
5
5
5
i2 = 3i =
3
A
5
&
3
v2 = +40i2 = +40 × = 24V
5
3
72
p40 Ω = +i2 v2 = + × 24 = W
5
5
4
96
p4iA = −(4i )v2 = −( ) × 24 = −
= −19.2W
5
5
∑p
dis
∑p
∴
gen
= 4+
4 72 20 + 4 + 72 96
+
=
=
= 19.2W
5 5
5
5
= 19.2W
∑p
dis
= ∑ pgen = 19.2W
Ω
Ω
Figure 12
(1)