Superposition

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# Superposition - PowerPoint PPT Presentation

Superposition. For a linear circuit with more than one independent source, superposition holds, i.e. any output voltage or current can be calculated as the sum of the contributions due to each source acting alone. Drill: #5 p. 158. 3A. 2 W. +. 2 W. 2 W. V o. 1A. 14V. +. Linear

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Presentation Transcript
Superposition
• For a linear circuit with more than one independent source, superposition holds, i.e. any output voltage or current can be calculated as the sum of the contributions due to each source acting alone.
• Drill: #5 p. 158

3A

2 W

+

2 W

2 W

Vo

1A

14V

+

Linear

Resistive

Elements

vo

+

va

ib

+

Proportionality
• For a linear circuit when any independent sources is acting alone, any output voltage or current is proportional to that single source i.e. if the source is multiplied by K, the output is also multiplied by K.
• Drill: #2(a) p. 157: vo= 3v12+4v22
• Drill: #9 p.158:
• va=10V and ib=0  vo=5V
• va=0 and ib=10A  vo=1V
• va=20V and ib=20A  vo=?

+

+

Vo

R2

Superposition and Dependent Sources
• Method 1: Apply superposition to independent sources ONLY.
• Method 2: Apply superposition to dependent and independent sources
• Drill: #21 p. 160:
• Treat the dependent current source as an independent source Ix
• Find the output Vo and the control Vg by superposition:
• Ix = 0  Vo1=

Vg1 =

• Vi = 0  Vo2=

Vg2 =

• (1) Vo=Vo1+Vo2

(2) Vg = Vg1+Vg2

Solve (2) for Vg and plug into (1) 

gmVg

+

Vg

R1

Vin

+

i

R

i

i

+

+

+

v

v

v

+

R

R

Vs

Is

Is

Dependent sources

18V

2 W

3 W

5 W

2 W

Vs = IsR

Vs = IsR

+

+

1A

12V

Source Transformation Theorem
• A series connection of a voltage source, Vs, and a resistor, R, is equivalent to a parallel connection of a current source, Is=Vs/R, and a resistor R. (Equivalent  have the same i-v relationship)
• Drill: #22 p160: Apply source transformation theorem to find P(5W)

Since P= RI2 = V2/R then need I(5W) or V(5W)

R

i

+

v

Vs

Independent sources

+

i

i

Passive

Network

Rth

+

+

Vth

v

v

Thévenin Theorem for Passive Networks
• Passive network: circuit containing resistors and independent sources.
• Thévenin theorem for passive networks: Any two-terminal passive network can be reduced to an equivalent two-terminal network consisting of a voltage source, Vth, in series with a resistance, Rth.

i

i

Passive

Network

+

+

Rth

IN

v

v

Norton Theorem for Passive Networks
• Norton theorem for passive networks: Any two-terminal passive network can be reduced to an equivalent two-terminal network consisting of a current source, IN, in parallel with a resistance, Rth.
Finding the Thevenin /Norton Equivalent
• Method 1: use source transformation to reduced the circuit to a resistance in series/parallel with voltage/current source
• Drill: #27 p.160
• Method 2: calculate Voc , and find Rth=Req with all independent sources set to zero i.e. Vs= 0 and Is= 0.
• Drill: #28 p.161
• Method 3: calculate Voc and isc

Vth=Voc , IN=isc and Rth= Rth = Voc/isc

• Drill: #29 p. 161

+

+

+

v

Vth

i

isc

Finding Thevenin /Norton Equivalent (cont.)
• Method 4: i-v measurements
• Drill: #37 p.162

i

i

Rth

+

+

Vth

v

Rth

IN

v

Thévenin&Norton Theorems for Active Networks
• Linear active network: linear circuit containing resistors, independent and dependent sources.
• Thévenin theorem for linear active networks: almost any two-terminal linear active network can be reduced to a Thévenin equivalent.
• Norton theorem for active networks:almost any two-terminal linear active network can be reduced to a Norton equivalent.
• ONLY method 3 and 4can be used to find the Thévenin/Norton equivalent of a linear active network.
• Drill: #31 p. 162
• Drill: #35 p.162

+

+

Rs

RL

Vs

PL

Rs

RL

Maximum Power Transfer
• Practical voltage/current sources dissipate part of the power they deliver internally (they heat up).
• Internal power dissipation modeled by a resistor:
• Power delivered to a load, RL/ power dissipated by RL:

Rs

+

Rs

Vs

Is

Vs

Is

Ideal Sources

Practical Sources

6 W

+

3 W

RL

VL

3A

Maximum Power Transfer (Cont.)
• Maximum power transfer for RL = Rs 
• Drill: #45 p. 164