ABE425 Engineering Measurement
Systems
Circuit Analysis
Dr. Tony E. Grift
Dept. of Agricultural & Biological Engineering
University of Illinois
Agenda
Fundamental laws of electricity in circuits
Kirchoff’s Current Law (KCL)
Kirchoff’s Voltage Law (KVL)
Basic circuits
Series resistance
Parallel resistance
Voltage divider
Current divider
Circuit simplification
Thevenin equivalent
Norton equivalent
Fundamental laws of electricity in circuits
Kirchoff’s Current Law
In a node the algebraic sum of currents equals zero
i1
i3
i2
Kirchoff’s Voltage Law
In a loop the algebraic sum of voltages equals zero
i1  i2  i3  0
i1  i2  i3
The equivalent resistance
of two resistances in
series is the sum of the
resistances.

U1  U 
R1 

U1  U 2 
i

 R1  R2 
U U2 
i 
R2 
 REQ  R1  R2
i 


U1  U 2
REQ 

i

The equivalent resistance
of two resistances in
parallel is the product
divided by the sum of the
resistances.
KCL : i  i1  i2
U1  U 2 

i1 

R1 
 1 1 
 i1  i2  i  U1  U 2    
U1  U 2 
R1 R2  
R1 R2

i2 
REQ 


R2 
R1  R2


U1  U 2
i

REQ

If you have a voltage divider and you
want to know the voltage drop across
one resistor, you take that resistor
itself, you divide by the sum of both
resistors and you multiply by the
voltage drop across both of them.
U1  U 2
R1  R2


R2

U1  U 2 
U OUT  U 2 
U OUT  U 2 
R1  R2

U
i
R2

R2
i
U OUT  U 2  U R2  U OUT  U 2  U R2
Let’s see how a
voltage divider works.
Let’s calculate the
current i to get all
voltages.
U3  U0
30  0
i

 0.5 A
R3  R2  R1 10  20  30
Now we can use Ohm’s
Law to calculate voltage
DROPS.
U R3  iR3  0.5*10  5V
U R2  iR2  0.5* 20  10V
U R1  iR1  0.5*30  15V
Now we can calculate all
the voltages.
U1  0  U R1  15V
U 2  U1  U R2  15V  10V  25V
U 3  U 2  U R3  25V  5V  30V
Let’s verify the voltage
DROPS using the Voltage
Divider rule.
R1
10
U R1 
U 3  U1  
 30  15  5V
R1  R2
10  20
U R2
R2
20

U 3  U1  
 30  15  10V
R2  R1
10  20
U R3
R3
30

U 2  0  
 25  0   15V
R3  R2
30  20
If you have a current divider and you
want to know the current through one
resistor, you take the opposite
resistor, you divide by the sum of
both resistors and you multiply by the
current into both of them.
U1  U 2 
 R1 R2 

 R1 R2 



R

R
R2



1
2 
i
i
 R1  R2   i1 
R1
R1  R2

U U2

i1  1
R1

i
Thevenin equivalent circuit
• Thevenin: replace a complicated circuit with a simple voltage
source with a series resistance
Thevenin Equivalent
Thevenin equivalent circuit procedure
• Step 1: Identify the circuit of which the Thevenin
equivalent is to be determined
• Step 2: Determine the Open Voltage across the
terminals of interest
• Step 3: Turn OFF voltage sources, (short voltage
sources, open-circuit all current sources and
determine the equivalent series resistance that a
voltage source at the load location would see
• Step 4: Replace the circuit with its equivalent Voltage
source and the equivalent series resistance
computed in steps 2 and 3
Thevenin equivalent circuit procedure
• Step 1: Identify the circuit of which the Thevenin
equivalent is to be determined
Thevenin equivalent circuit procedure
• Step 2: Determine the Open Voltage across the
terminals of interest
Thevenin equivalent circuit procedure
• Step 3: Turn OFF voltage sources, (short voltage
sources, open-circuit all current sources) and
determine the equivalent series resistance that a
voltage source at the load location would see
Thevenin equivalent circuit procedure
• Step 4: Replace the circuit with its equivalent Voltage
source and the equivalent series resistance
computed in steps 2 and 3
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1) Compute the Open Voltage
Va ,b
R1
a
+
VS
-
Va ,b
R2
iS
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1A) Contribution of Voltage source VS
• Remember that the internal resistance of the current
source is infinite!
R2
Va ,b V 
VS
S
R1  R2
R1
a
+
VS
-
Va ,b
R2
iS
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1B) Contribution of Current source iS
• Remember that the internal resistance of the Voltage
source is 0!
R1
a
+
VS
-
Va ,b
R2
iS
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1B) Contribution of Current source iS
• Remember that the internal resistance of the Voltage
source is 0!
R1 R2
Va ,b i  iS  R1 / / R2   iS
S
R1  R2
R1
a
+
VS
-
Va ,b
R2
iS
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1C) Add contributions of both sources U , i
• In linear circuits this is allowed and called superposition
R1
a
+
U
-
R2
U a ,b
i
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1C) Add contributions of both sources U , i
• In linear circuits this is allowed and called superposition
R2
RR
R2
Va ,b  Va ,b V  Va ,b i  VS
 iS 1 2 
VS  iS R1 

S
S
R1  R2
R1  R2 R1  R2
Va ,b  Va ,b V  Va ,b i
S
S
3

10  5*7   13.5V
73
R1
a
+
U
-
R2
U a ,b
i
b
Thevenin equivalent circuit procedure
• 2A) Compute the equivalent series resistance that a
voltage source at the load location a,b would see
R1
a
R2
b
Thevenin equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 2A) Compute the equivalent series resistance that a
voltage source at the load location a,b would see
R1R2
7 *3
Reqv  R1 // R2 

 2.1
R1  R2 7  3
R1
a
R2
b
Thevenin equivalent circuit procedure
• Step 4: Replace the circuit with its equivalent Voltage
source and the equivalent series resistance
computed in steps 2 and 3
Va ,b  13.5V
Reqv  2.1
Norton equivalent circuit
• Norton: replace a complicated circuit with a simple
current source with a parallel resistance
Norton Equivalent
Norton equivalent circuit procedure
• Step 1: Identify the circuit of which the Norton
equivalent is to be determined
• Step 2: Determine the short current across the
terminals of interest
• Step 3: Turn OFF voltage sources, (short voltage
sources, open-circuit all current sources and
determine the equivalent parallel resistance that a
voltage source at the load location would see
• Step 4: Replace the circuit with its equivalent Current
source and the equivalent parallel resistance
computed in steps 2 and 3
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1) Compute the Short Current through load
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1A) Contribution of Voltage source V S
• Remember that the internal resistance of the current
source is infinite!
iN
VS
VS

R1
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1B) Contribution of Current source i S
• Current will take the path of least resistance!
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1C) Contribution of Current source i S
• Current will take the path of least resistance!
iN
iS
 is
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1D) Add contributions of both sources VS , iS
• In linear circuits this is allowed and called superposition
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 1D) Add contributions of both sources VS , iS
• In linear circuits this is allowed and called superposition
iN
VS
 iN
iS
VS
10
45

 iS   5 
R1
7
7
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 2) Compute the equivalent parallel resistance that a
current source at the load location would see
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 2) Compute the equivalent parallel resistance that a
current source at the load location would see
R1R2
7 *3
Reqv  R1 // R2 

 2.1
R1  R2 7  3
Norton equivalent circuit procedure
VS  10V , R1  7, R2  3, iS  5 A
• 2) Compute the equivalent parallel resistance that a
current source at the load location would see
iN 
VS
10
45
 iS   3 
R1
7
7
Reqv 
R1 * R2 7 * 3

 2.1
R1  R2 7  3
Compare Thevenin and Norton procedures
VS  10V , R1  7, R2  3, iS  5 A
VT  13.5V 

45
45

iN 
A 3 * 4.5 
* 3 * 0 .7
7
7

Reqv  2.1 
VT  iN * Reqv
ABE425 Engineering Measurement
Systems
Circuit Analysis
The End
Dept. of Agricultural & Biological Engineering
University of Illinois