Basic Circuit Theorems, Laws
and Techniques
Dr. : Mohamed Hassan Abd El-azeem
10/2/2013
Dr. MOHAMED HASSAN
Lecture 2
Contents:
1. Part 1 : Ohm’s Law
Kirchhoff’s Laws
Voltage Divider
Current Divider
2. Part 2 : Source Transformation
2.1 Ohm’s Law
 Current and voltage sources in
action
 Compute current across R
vs = 5 V
+
-
1 kW
v
5V
i   3  5mA
R 10 W
 Compute voltage across R
is = 7
mA
10/2/2013
1 kW
V  iR  7 mA 1kW  7V
Dr. MOHAMED HASSAN
2.2 Kirchhoff’s Laws
 Nodes and loops
 Nodes are points were 2
or more elements meet
v1
+
-
R1
R2
R3
10/2/2013
 Loops are closed paths
around the circuit
starting and ending at a
node
Dr. MOHAMED HASSAN
2.2.1 Kirchhoff’s Current Law
 Charges entering a node are exactly the
same as ones leaving the node. Sum of
currents at a node equal 0.
qt
qt
qt
qt
qt
qt
qt
10/2/2013
Dr. MOHAMED HASSAN
2.2.1 Kirchhoff’s Current Law
 Circuit Examplei
1
i2
i3
i4
is
is = i1 + i2 + i3 + i4
10/2/2013
is - i1 - i2 - i3 - i4 = 0
Dr. MOHAMED HASSAN
2.2.2 Kirchhoff’s Voltage Law
 Sum of voltages around a loop equal 0
i1
v1
+
-
i2
+
-
+ R1
i3
R4
+
R2
-
-
R3
+
Loop 1: -v1 + R1i2 + R2i2 + R3i2 = 0
Loop 2: -v1 + R4i3 = 0
Loop 3: - R4i3 + R1i2 + R2i2 + R3i2 = 0
10/2/2013
Dr. MOHAMED HASSAN
2.3 Kirchhoff’s Laws Exercise
 Compute i1, i2, & i3, & p expended for all resistors.
Given R1 = R2 = R3 = 10kW, R4 = 15kW, & v1 = 15V
Loop Equations
i2
i1
-v1 + i2(R1 + R2 + R3) = 0
-v1 + R4i3 = 0
- R4i3 + R1i2 + R2i2 + R3i2 = 0
v1
+
-
+
+ R
1
-
i1 - i2 - i3 = 0
-15 + 15e3 i3 = 0
-v1 + i2(R1 + R2 + R3) = 0
10/2/2013
R2
-
R3
+
i3 = 1mA
-15 + i2(30e3) = 0
Dr. MOHAMED HASSAN
+
-
Node Equation(s)
-v1 + R4i3 = 0
i3
R4
-
i2 = 0.5mA
2.3 Kirchhoff’s Laws Exercise
 Given R1 = R2 = R3 = 10kW, R4 = 15kW, & v1 =
Loop 15V
Equations
i2
i1
-v1 + i2(R1 + R2 + R3) = 0
-v1 + R4i3 = 0
- R4i3 + R1i2 + R2i2 + R3i2 = 0
+
v1
+
-
+ R1
i3
R4
Node Equation(s)
R2
-
-
i3 = 1mA i1 - i2 - i3 = 0
R3
i1 = 1.5mA
pR1 = pR2 = pR3 = (i2)2R= (0.5mA x 0.5mA)10e3=2.5mW
pR1 = (i3)2R= (1mA)215e3=15mW
10/2/2013
+
-
i1 - i2 - i3 = 0
i2 = 0.5mA
-
Dr. MOHAMED HASSAN
+
2.4 Voltage Divider
is
vs
is 
R1  R2
R1
v1 
vs
R1  R2
+
R1
2 October 2013
-
+
vs
+
-
R2
R2
v2 
vs
R1  R2
v1
v2
-
Dr. MOHAMED HASSAN
Voltage Divider
 Find vo in terms of vs
is
R L R2
Req 
R L  R2
R1
+
vs
+
-
R2
vo
-
2 October 2013
RL
Vo 
Req
R1  Req
Vs
R L R2
R L  R2
Vo 
Vs
R L R2
R1 
R L  R2
R L R2

Vs
R1  R L  R2   R L R2
Dr. MOHAMED HASSAN
2.5 Current Divider
 Find i1 and i2 in terms of is
i1 R1  i2 R2
is  i1  i2
 R2  R1 
R1
R1

i2  i1
 is  i1  i1
 i1 
R2
R2
 R2 
 R2
i1  is 
 R1  R2



 R1
i2  is 
 R1  R2



2 October 2013
is
Dr. MOHAMED HASSAN
R1
i1
R2
i2
Note:
Resistors in Series & Parallel
 Find equivalent resistance
5kW
5kW
35kW
10V
10V
+
-
+
100kW
5kW
25kW
100kW
-
40kW
5kW
5kW
10V
10V
+
-
50kW
5kW
10/2/2013
100kW
Dr. MOHAMED HASSAN
+
-
60kW
Example 2-1
i1
Find v1 and v2 in the circuit shown in
Fig., Also calculate i1 and i2
and the power dissipated in the 12-Ω and
40-Ω resistors.
12 Ω
+ V1 -
6Ω
i2
Solution:-
+
12*6
=4W
12+6
12
6=
40
40*10
10 =
=8W
40+10
V1 = 15
4
4+8
=5V
V2 = 15
8
4+8
= 10 V
V1
5V
=
= 416.7 mA
12 W
12 W
V2
10 V
i2 =
=
= 250 mA
40 W
40 W
P12W =  i1 
P40W =  i2 
2
12 =
40 =
 V1 
4Ω
+ V1 -
2
12
2
 V2 
40
V2 40 Ω
-
i1 =
2
10 Ω
15 V +-
= 2.083 W
= 2.5 W
+
15 V +-
V2 8 Ω
-
Example 2.2
For the circuit shown in Fig., find:
(a) v1 and v2, (b) the power dissipated
in the 3-k and 20-k resistors, and
(c) the power supplied by the current
source.
Solution:i1 = iS
R eq
i2 = iS
4 kW
R eq
20 k W
i1 =10 mA
R eq = 2 kΩ
2
3 kW =
 V1 
2
3 kW
+
3 kΩ
V
1
-
= 75 mW
+
i1
10 mA
i2
V2 20 kΩ
5 kΩ
-
1
1
1
1
=
+
+
R eq
20 k W
5 kW
4 kW
2 kW
= 5 mA
4 kW
V1 = i1 3 kW = 1 5 V
P3-k =  i1 
1 kΩ
i2 =10 mA
2 kW
= 1 mA
20 k W
V2 = i2 20 kW = 2 0 V
P20-k =  i2 
2
20 k W =
P10-mA = - iS * V2 = - 10 mA * 20 V = - 200 mW
 V2 
2
20 k W
= 20 m W
2.6 Source Transformation
 We can convert a voltage source with series resistor
to a current source with parallel resistor.
If we place a load then the voltage across the load should be the same
R
+
+
is R
+
v
L
-
-
v
-
vs=isR
2 October 2013
vs
Dr. MOHAMED HASSAN
Source Transformation
 More simplications
is R
is R
R
vs
+
-
2 October 2013
R
vs
+
Dr. MOHAMED HASSAN
-
Source Transformation
Example
 Find v0
25W
8A
250V
+
125W
-
+
v0
100W
20W
10W
-
25W
250V
8A
+
+
v0
-
20W
100W
2 October 2013
8A
10A
25W
+
v0
-
Dr. MOHAMED HASSAN
20W
100W
2.7 Source Transformation
Example : Find V0
8A
10A
25W
+
v0
+
20W
2A
100W
25W
-
-
+
2A
v0
10W
-
2 October 2013
v0
Dr. MOHAMED HASSAN
v0= 20V
20W
100W
2Ω
Example 2-3
3Ω
+
In the circuit shown in Fig., use the ST to find
4Ω
vO
8Ω
3A
4Ω
12 V, 4-Ω (Series)
12 V, 3-Ω (Series)
4 A, 3-Ω (║)
4Ω+2Ω=6Ω
12 V, 6-Ω (Series)
2Ω
+
+
12 V
8Ω
6 Ω ║ 3 Ω = 6 * 3 / (6+3) = 2 Ω
4 A (up) + 2 A (down)
i=
3Ω
4A
3Ω
4A
+
2A
6Ω
i
vO = 8 * i = 8 * 0.4 = 3.2 V
8Ω
Dr. MOHAMED HASSAN
8Ω
Vo
-
2 A (up)
2
(2 A) = 0.4 A
2+8
2 October 2013
Vo
-
2 A, 6-Ω (║)
Also, current sources in parallel can be
(combined) added together according to
the current direction, in our case here
12 V +-
-
Solution
3 A, 4-Ω (║)
Vo
+
Vo
-
2Ω
2A
Example 2.4
6Ω
4Ω
5Ω
In the circuit shown in Fig., calculate P6V
Solution
6 V +-
40 V, 5-Ω (Series)
(║)
20 Ω
30 Ω
+ 40 V
-
8 A, 5-Ω
4Ω
5 Ω ║ 20 Ω = 5 * 20 / (5+20) = 4 Ω
8 A, 4-Ω (║)
(Series)
10 Ω
6Ω
32 V, 4-Ω
6 V +-
20 Ω
30 Ω
5Ω
8A
6 Ω + 4 Ω + 10 Ω = 20 Ω
32 V, 20-Ω (Series)
(║)
1.6 A, 20-Ω
4Ω
10 Ω
6Ω
4Ω
30 Ω ║ 20 Ω = 30 * 20 / (30+20) = 12 Ω
1.6 A, 12-Ω (║)
(Series)
6 V +-
19.2 V, 12-Ω
i = 19.2 – 6 / (12 + 4) = 0.825 A
4Ω
+
-
30 Ω
32 V
10 Ω
4Ω
12 Ω
P6V = + (6 V) * (0.825 A) = 4.95 W
i
6 V +-
(Delivered)
2 October 2013
+
-
19.2 V
6 V +Dr. MOHAMED HASSAN
30 Ω
20 Ω
1.6 A
Example 2.5
1.6 Ω
In the circuit shown in Fig., use the ST to
20 Ω
calculate (a) V and (b) P120V
+
+
60 V
36 A
Solution
+ 120 V
-
6Ω
V
5Ω
8Ω
-
(a)
120 V, 20-Ω (Series)
6 A, 20-Ω (║)
1.6 Ω
60 V, 5-Ω (Series)
+
12 A, 5-Ω (║)
6A
20 Ω
12A
5Ω
36 A
6Ω
V
20 Ω ║ 5 Ω ║ 6 Ω = 2.4 Ω
8Ω
-
6A (up) + 12A (down) + 36A (up)
30 (up)
30 A, 2.4-Ω (║)
2.4 Ω
+
72 V, 2.4-Ω (Series)
2 October 2013
72 V +-
72
V=
(8) = 48 V
2.4 + 1.6 + 8
1.6 Ω
V
-
Dr. MOHAMED HASSAN
8Ω
(b)
1.6 Ω
60 V, 5-Ω (Series)
20 Ω
12 A, 5-Ω (║)
5 Ω ║ 6 Ω ║ (8 Ω + 1.6 Ω) = 2.124 Ω
12A
+
-
5Ω
36 A
6Ω
8Ω
120 V
36A (up) + 12A (down) = 24 (up)
24 A, 2.124-Ω (║)
50.7 V, 2.124-Ω (Series)
20 Ω
i=
120 - 50.7
= 3.12 A
20 + 2.124
P120V = - 120 (3.12) = 374.4 W
120 V +-
(Delivered)
20 Ω
120 V +-
2 October 2013
2.124 Ω
24A
Dr. MOHAMED HASSAN
i
2.124 Ω
+ 50.7 V
-
Example 2.6
In the circuit shown in Fig., use
5V
-+
the ST to calculate iO
Answer
iO = 1.78 A
2 October 2013
6Ω
5A
3Ω
Dr. MOHAMED HASSAN
7Ω
1Ω
i
o
3A
4Ω