1
|
|
2
|
- Voltage across (or current through) an element
- Determined by summing voltage (or current) due to each independent
source
- All sources (except dependent sources) other than the one being
considered are eliminated
|
3
|
- Replace current sources with opens
- Replace voltage sources with shorts
|
4
|
- Circuit may operate at more than one frequency at a time
- Superposition is the only analysis method that can be used in this case
- Reactances must be recalculated for each different frequency
|
5
|
- Diode and transistor circuits will have both dc and ac sources
- Superposition can still be applied
|
6
|
- Superposition theorem can be applied only to voltage and current
- It cannot be used to solve for total power dissipated by an element
- Power is not a linear quantity
- Follows a square-law relationship
|
7
|
- If controlling element is external to the circuit under consideration
- Method is the same as for independent sources
|
8
|
- Simply remove sources one at a time and solve for desired voltage or
current
- Combine the results
|
9
|
- If the dependent source is controlled by an element located in the
circuit
- Analysis is different
- Dependent source cannot be eliminated
|
10
|
- Circuit must be analyzed by considering all effects simultaneously
|
11
|
- Converts an ac circuit into a single ac voltage source in series with an
equivalent impedance
- First, identify and remove the element or elements across which the
equivalent circuit is to be found
|
12
|
- Label two open terminals
- Set all sources to zero
- Replace voltage sources with shorts
- Current sources with opens
|
13
|
- Calculate the Thévenin equivalent impedance
- Replace the sources and determine
open-circuit voltage
|
14
|
- If more than one source is involved
- Superposition may be used
- Draw resulting Thévenin equivalent circuit
- Including the portion removed
|
15
|
- Converts an ac network into an equivalent circuit
- Consists of a single current source and a parallel impedance
- First, identify and remove the element or elements across which the
Norton circuit is to be found
|
16
|
- Label the open terminals
- Set all sources to zero
|
17
|
- Determine Norton equivalent impedance
- Replace sources and calculate short-circuit current
|
18
|
- Superposition may be used for multiple sources
- Draw resulting Norton circuit
- Including portion removed
|
19
|
- Possible to find Norton equivalent circuit from Thévenin equivalent
circuit
- Use source transformation method
- ZN = ZTh
- IN = ETh/ZTh
|
20
|
- If a circuit contains a dependent source controlled by an element
outside the area of interest
- Previous methods can be used to find the Thévenin or Norton circuit
|
21
|
- If a circuit contains a dependent source controlled by an element in the
circuit
- Other methods must be used
|
22
|
- If a circuit has a dependent source controlled by an element in the
circuit
- Use following steps to determine equivalent circuit
|
23
|
- First
- Identify and remove branch across equivalent circuit is to be
determined
- Label the open terminals
|
24
|
- Calculate open-circuit voltage
- Dependent source cannot be set to zero
- Its effects must be considered
- Determine the short-circuit current
|
25
|
- ZN = ZTh = ETh/IN
- Draw equivalent circuit, replacing the removed branch
|
26
|
- A circuit may have more than one independent source
- It is necessary to determine the open-circuit voltage and short-circuit
current due to each independent source
|
27
|
- Effects of dependent source must be considered simultaneously
|
28
|
- Maximum power
- Delivered to a load when the load impedance is the complex conjugate of
the Thévenin or Norton impedance
|
29
|
- ZTh = 3W + j4W
ZL = ZTh*
= 3W - j4W
- ZTh = 10 WÐ30° ZL = ZTh*
= 10 WÐ-30°
|
30
|
- If the ZL is the complex conjugate of ZTh or ZN
|
31
|
- If it is not possible to adjust reactance part of a load
- A relative maximum power will be delivered
- Load resistance has a value determined by
|