ECET 2111 Circuits II Network Theorems
Superposition Theorem Superposition Theorem: Given a linear network that contains multiple sources…
Any voltage (or current) in the network may be determined by solving for that voltage (or current) with each of the sources individually “turned-on” (and all of the other sources “turned-off ”), and then summing the individual solutions.
Z1 Z2 +
+ ~ ~ ~ VA V3 Z3 IB
-
1 Ideal Sources “Turned-Off” Since an Ideal Voltage Source maintains a constant voltage independent of current flow, if the source is “turned off” (set to zero volts), then the voltage across the source’s terminals will be zero independent of the source’s current.
This is equivalent to the characteristics of an “ideal wire”…
Thus, an Ideal Voltage Source that is “turned-off” acts like an ideal wire: ~ ~ I I + ~ ~ V = 0 V = 0
Ideal Sources “Turned-Off” Since an Ideal Current Source maintains a constant current independent of the source voltage, if the source is “turned off” (set to zero volts), then the current flowing through the source will be zero independent of the voltage across its terminals.
This is equivalent to the characteristics of an “open-circuit”…
Thus, an Ideal Current Source that is “turned-off” acts like an open-circuit: ~ I = 0 ~ ~ ~ V I = 0 V
2 Superposition Theorem Example Superposition Theorem Example: Given the following circuit containing two sources…
Determine the voltage V3 using the superposition theorem.
Z1 Z2 200 + 80 j60 ~ + ~ VA ~ IB V3 Z3 j200 1000 V 1.230 A
-
Superposition Theorem Example
In order to determine the voltage V3 using the superposition theorem, the circuit must be analyzed with each of the sources individually “turned-on” one at a time:
A1 B1 Z1 Z2 Z1 Z2 200 + 80 j60 200 + 80 j60 ~ + ~ VA ~ ~ IB V3A Z3 j200 V3B Z3 j200 1000 V 1.230 A
- -
A2 B2 Source A “turned-on” Source B “turned-on”
3 Superposition Theorem Example After the contributions of the individually-sourced circuits to the voltage V3 are determined, the results can be summed in order to determine its actual value. ~ ~ ~ V3 V3A V3B
A1 B1 Z1 Z2 Z1 Z2 200 + 80 j60 200 + 80 j60 ~ + ~ VA ~ ~ IB V3A Z3 j200 V3B Z3 j200 1000 V 1.230 A
- -
A2 B2 Source A “turned-on” Source B “turned-on”
Superposition Theorem Example
Solve for V3A: Since a current source reacts like an open-circuit when it is turned-off, the only current path is through Z1 and Z3, allowing V3A to be determined using a voltage divider: ~ j200 V3A 1000 70.7 45 V 200 j200
A1 A1 Z1 Z2 Z1 200 + 80 j60 200 + ~ + ~ + VA ~ VA ~ V3A Z3 j200 Z3 j200 1000 V 1000 V V3A
- -
A2 A2 Source A “turned-on”
4 Superposition Theorem Example
Solve for V3B: Since a voltage source reacts like a short-circuit when it is turned-off, source “A” can be replaced by an ideal wire, resulting in the parallel connection of Z1 and Z3. 1 1 1 200 j200 100 j100 200 j200
B1 B1 Z1 Z2 Z2 200 + 80 j60 + 80 j60 ~ ~ ~ IB ~ IB V3B Z3 j200 V3B Z13 100 j100 1.230 A 1.230 A
- -
B2 B2 Source B “turned-on”
Superposition Theorem Example
Solve for V3B:
Once Z1 and Z3 are replaced by the parallel-equivalent Z13, V3B can be determined using Ohm’s Law: ~ V3B 1.230100 j100 169.7 15 V
B1 B1 Z1 Z2 Z2 200 + 80 j60 + 80 j60 ~ ~ ~ IB ~ IB V3B Z3 j200 V3B Z13 100 j100 1.230 A 1.230 A
- -
B2 B2 Source B “turned-on”
5 Superposition Theorem Example
Determine V3:
Now that V3A and V3B have been solved, the actual voltage V3 can be determined by summing the individual results: ~ ~ ~ V3 V3A V3B 70.7 45 169.7 15 233.6 23.7 V
Z1 Z2 200 + 80 j60 ~ + ~ VA ~ IB V3 Z3 j200 1000 V 1.230 A
-
Superposition Theorem Example Check the Superposition results using Nodal Analysis:
If the bottom node is chosen as ground and the top node is assigned the node-voltage variable Vx , then the voltage rise from ground to node “x” is equivalent to V3.
~ Vx Z1 Z2 200 + 80 j60
+ ~ 1000 V V3 Z3 j200 1.230 A
-
6 Superposition Theorem Example Check the Superposition results using Nodal Analysis :
Write the KVL equation for node “x”: ~ ~ ~ I1 I2 I3 0 ~ ~ V 1000 V 0 x 1.230 x 0 200 j200
~ Vx Z1 ~ ~ Z2 200 I1 I2 80 j60 ~ + I3 ~ 1000 V V3 Z3 j200 1.230 A
Superposition Theorem Example Check the Superposition results using Nodal Analysis : ~ ~ V 1000 V 0 Determine V from the node equation: x 1.230 x 0 x 200 j200 ~ ~ V 1000 V x x 1.230 200 200 j200 ~ 1 1 1000 Vx 1.230 200 j200 200 ~ Vx 0.005 j0.005 1.652 21.29
~ 1.652 21.29 V 0.005 j0.005 223.6 23.7 V x 0.005 j0.005
7 Thevenin’s Theorem
Thevenin’s Theorem Thevenin’s theorem provides a method for modeling the response of a complex single-port network by replacing the complex network with a simple “equivalent” single-port network that will have the same response as the original network to any externally-connected devices.
T
Single-Port External Network Devices Connected Here H
8 Thevenin’s Theorem Thevenin’s theorem: (For steady-state AC circuits) Any linear, single-port network can be replaced by a simple equivalent network that consists of a single voltage source, VTH , in series with a single impedance, ZTH , without affecting the operation of any devices that are externally connected to the network’s terminals.
T ZTH T
Single-Port External + Network Devices ~ VTH Connected Here H H
Thevenin’s Theorem The simple equivalent network is often referred to as Thevenin’s Equivalent Circuit, where:
• VTH is Thevenin’s Equivalent Voltage, and
• ZTH is Thevenin’s Equivalent Impedance.
T ZTH A
Single-Port External + Network Devices ~ VTH Connected Here B H
9 Thevenin’s Theorem
The values for VTH and ZTH can be determined as follows:
The value for VTH is obtained by determining the open- circuited terminal voltage of the single-port network with all sources in the network “turned-on”.
The value for ZTH is obtained by determining the resistance “seen” looking into the terminals of the original single- port network with all of the independent sources contained within the network “turned-off”.
Maximum Power Transfer Theorem
10 Maximum Power Transfer Theorem The Maximum Power Transfer Theorem states that, in order to transfer maximum power from a linear, single-port, AC network to a single, arbitrary impedance-type load where:
Z L RL jX L the value of the impedance should be set equal to the complex conjugate of the single-port network’s Thevenin Equivalent Impedance.
If: ZTH RTH jX TH T Single-Port Network * ZL Then set: Z L ZTH RTH jX TH H
Maximum Power Transfer Theorem Additionally, the Maximum Power Transfer Theorem states that, in order to transfer maximum power from the single- port network to an arbitrary, purely-resistive load where:
Z L RL the value of the load resistance should be set equal to the magnitude of the Thevenin Equivalent Impedance.
If: ZTH RTH jX TH T Single-Port Network 2 2 ZL Then set: RL ZTH RTH X TH H
11 Thevenin’s Theorem Example Determine the Thevenin’s Equivalent Circuit for the following network with respect to terminals “T” and “H”:
A T 40 26 RTH
+ + 25V 150 100 VTH
H B
Thevenin’s Theorem Example
Determine Thevenin’s Resistance, RTH:
T The value for R is obtained 40 26 TH
+ by determining the resistance off 150 100 RTH looking into terminals “T” and “H” while all sources are H “turned-off”.
12 Thevenin’s Theorem Example
Determine Thevenin’s Resistance, RTH:
T The value for R is obtained 40 26 TH
+ by determining the resistance off 150 100 RTH looking into terminals “T” and “H” while all sources are H “turned-off”.
st T 1 – “Turn-off” all sources. 40 26 Remember that an ideal off 150 100 RTH voltage source, when “turned-off”, looks like H an ideal wire.
Thevenin’s Theorem Example
Determine Thevenin’s Resistance, RTH:
nd T 2 – Place a “test” source across 40 26 terminals “T” and “H”, and 150 100 RTH reduce the network connected to the “test” source down to a H single equivalent resistance.
T Replace 40||150||100 with 26 single equivalent resistance:
24 RTH 1 1 1 1
H Req 24 40 150 100
13 Thevenin’s Theorem Example
Determine Thevenin’s Resistance, RTH:
T Finally – Replace the 24 and 26 26 series-connected resistors 24 RTH with a single equivalent resistance: H
Req 24 26 50
T The resistance seen “looking” into terminals “T” and “H” of 50 RTH the network with all sources “turned-off” is: H
RTH 50
Thevenin’s Theorem Example
Determine Thevenin’s Voltage, VTH:
T + The value for V is obtained 40 26 TH
+ by determining the voltage 25V 150 100 VTH across terminals “T” and “H” while the terminals are open- H - circuited.
+ VY -
T + Based on KVL, VTH can be 40 26 + defined in terms of the resistor + 25V 150 100 VX VTH voltages VX and VY : - H - VTH VX VY
14 Thevenin’s Theorem Example
Determine Thevenin’s Voltage, VTH:
+ VY -
T + Since I must equal zero due 40 26 Y + IY=0 + to the open-circuit between 25V 150 100 VX VTH terminals “T” and “H”, then - VY must also equal zero, and: H -
VTH VX
Thus, the Thevenin’s Voltage can be determined by solving for the voltage, VX, across the 100 resistor.
Thevenin’s Theorem Example
Determine Thevenin’s Voltage, VTH:
st T + 1 – Reduce the circuit by 40 26 + + replacing the 150||100 25V 150 100 VX VTH resistors with a single - equivalent resistance. H - 1 1 1 Req 60 150 100 T + 40 26 + + Note that the voltage across the 25V 60 VX VTH 60 resistor will equal VX in - H - the previous circuit.
15 Thevenin’s Theorem Example
Determine Thevenin’s Voltage, VTH:
T + Since no current can flow 40 26 + IY=0 + through the 26 resistor, all 25V 60 VX VTH of the current that flows - through the 40 resistor must - H flow through the 60 resistor.
Thus, the two resistors are operationally connected in-series, allowing VX to be solved using a voltage divider: 60 VTH VX 25 15V 40 60
Thevenin’s Theorem Example The Thevenin’s Equivalent Circuit for the following network with respect to terminals “T” and “H” is:
A T 40 26 RTH = 50
+ + 25V 150 100 VTH = 15V
H B
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