Circuits with Resistor
Combinations (2.6, 8.7)
Dr. Holbert
February 8, 2006
ECE201 Lect-7
1
Solving Circuits with Series and
Parallel Combinations
• The combination of series and parallel
impedances can be used to find voltages and
currents in circuits.
• This process can often yield the fastest
solutions to networks.
• This process may not apply to complicated
networks.
ECE201 Lect-7
2
Series and Parallel Impedances
• Impedances are combined to create a simple
circuit (usually one source and one
impedance), from which a voltage or
current can be found
• Once the voltage or current is found, KCL
and KVL are used to work back through the
network to find voltages and currents.
ECE201 Lect-7
3
Example: Resistor Ladder
1kW
1kW
+
10V
+
–
V1
1kW
+
2kW
V2
–
+
2kW
–
V3
1kW
–
Find V1, V2, and V3
ECE201 Lect-7
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Example: Resistor Ladder
1kW
1kW
+
10V
+
–
V1
1kW
+
2kW
V2
–
+
2kW
–
V3
1kW
–
Find an equivalent resistance for the network
with V1 across it, then find V1 using a
voltage divider.
ECE201 Lect-7
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Example: Resistor Ladder
1kW
+
10V
+
–
V1
1kW
–
1kW
V1  10V
 5V
1kW  1kW
ECE201 Lect-7
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Example: Resistor Ladder
1kW
10V
+
–
1kW
+
5V
–
1kW
+
V2
2kW
+
2kW
–
V3
1kW
–
Find an equivalent resistance for the network
with V2 across it, then find V2.
ECE201 Lect-7
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Example: Resistor Ladder
1kW
10V
+
–
1kW
+
5V
+
V2
2kW
–
1kW
–
1kW
V2  5V
 2.5V
1kW  1kW
ECE201 Lect-7
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Example: Resistor Ladder
1kW
10V
+
–
1kW
+
5V
–
+
2.5V
2kW
–
1kW
+
2kW
V3
1kW
–
1kW
V3  2.5V
 1.25V
1kW  1kW
ECE201 Lect-7
9
Example: Notch Filter
0.1W
70.4mH
100W
+
10V  0
+
–
100mF
Vout
1kW
–
Find Vout
Use w = 1500
ECE201 Lect-7
10
Example: Notch Filter
0.1W
j106W
100W
+
10V  0
+
–
–j6.67W
Vout
1kW
–
Find the equivalent impedance of the resistor,
inductor, and capacitor.
ECE201 Lect-7
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Example: Notch Filter
100W
7.12W  –89.99
+
10V  0
+
–
1kW
Vout
–
Combine top resistor and impedance.
ECE201 Lect-7
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Example: Notch Filter
100.3W  –4.07
+
10V  0
+
–
Vout
1kW
–
Vout
1kW0
 10V0
100.3W  4.07  1kW0
 9.09V  0.37
ECE201 Lect-7
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Example #2: Notch Filter
0.1W
70.4mH
100W
+
10V  0
+
–
100mF
Vout
1kW
–
Find Vout
Use w = 377
ECE201 Lect-7
14
Example: Notch Filter
Vout= 1.23V  0.17
ECE201 Lect-7
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Frequency Response
Magnitude of Vout
10
8
6
4
2
0
250
300
350
400
450
500
Frequency (rad/sec)
ECE201 Lect-7
16
Using MATLAB to Solve
Circuits
• MATLAB can perform computations with
complex numbers.
• You can use it as a calculator to compute
phasors and impedances for AC steady-state
analysis.
• You can also use it to automate
computations of frequency responses.
ECE201 Lect-7
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Using MATLAB
• Entering a complex number:
>> 1+2j
ans =
1.0000 + 2.0000i
• Multiplying complex numbers:
>> (1+2j)*(3+4j)
ans =
-5.0000 +10.0000i
ECE201 Lect-7
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Example: Notch Filter
0.1W
70.4mH
100W
+
10V  0
+
–
100mF
Vout
1kW
–
Find Vout
Use w = 1500
ECE201 Lect-7
19
Compute Impedances
>> omega = 377
omega =
377
>> xl = j*omega*70.4e-3
xl =
0 +26.5408i
>> xc = 1/(j*omega*100e-6)
xc =
0 -26.5252i
ECE201 Lect-7
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Equivalent Capacitor/Inductor
Impedance
>> zeq = (0.1+xl)*xc
/(0.1+xl+xc)
zeq =
6.8687e+03 - 1.0981e+03i
ECE201 Lect-7
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Voltage Divider
>> vin = 10
vin =
10
>> vout = vin*1e3
/(100+zeq+1e3)
vout =
1.2315 + 0.1697i
ECE201 Lect-7
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Magnitude and Angle
>> abs(vout)
ans =
1.2432
>> angle(vout)
ans =
0.1369
>> angle(vout)*180/pi
ans =
7.8461
ECE201 Lect-7
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Class Examples
• Learning Extension E2.14
• Learning Extension E2.15
• Learning Extension E8.12
ECE201 Lect-7
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