sinusoidal steady state analysis (ac analysis) part ii sinusoidal steady state analysis (ac...
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Sinusoidal Steady State Analysis
(AC Analysis)
Part II
Dr. Mohamed Refky Amin
Electronics and Electrical Communications Engineering Department (EECE)
Cairo University
http://scholar.cu.edu.eg/refky/
OUTLINE
• Previously on ELCN102
• AC Power Analysis
Instantaneous Power
Average Power
Complex Power
• Root Mean Square
• Maximum Average Power Transfer
Dr. Mohamed Refky 2
Previously on ELCN102
Dr. Mohamed Refky
Methods of Solution of AC CircuitsTo solve a AC circuit you can use one or more of the following
methods:
• Simplification Method
• Loop Analysis Method
• Node Analysis Method
• Superposition Method
• Thevenin equivalent circuit
• Norton equivalent circuit
Previously on ELCN102
Dr. Mohamed Refky
Simplification Method In step by step simplification we can use:
• Source transformation
• Combination of active elements
• Combination of series and parallel elements
• Star-delta & delta-star transformation
Previously on ELCN102
Dr. Mohamed Refky
Source Transformation“A voltage source 𝑉𝐴𝐶 with a series impedance 𝑍 can be
transformed into a current source 𝐼𝐴𝐶 = 𝑉𝐴𝐶/𝑍 and a parallel
impedance 𝑍”
“ A current source 𝐼𝐴𝐶 with a parallel impedance 𝑍 can be
transformed into a voltage source 𝑉𝐴𝐶 = 𝐼𝐴𝐶 × 𝑍 and a series
impedance 𝑍”
Previously on ELCN102
Dr. Mohamed Refky
Loop Analysis Method
The Loop Analysis Method (Mesh Method) uses KVL to generate
a set of simultaneous equations.
1) Convert the independent current sources into equivalent
voltage sources
2) Identify the number of independent loop (𝐿) on the circuit
3) Label a loop current on each loop.
4) Write an expression for the KVL around each loop.
5) Solve the simultaneous equations to get the loop currents.
Previously on ELCN102
Dr. Mohamed Refky
Matrix Form
𝑍11 −𝑍12 ⋯ −𝑍1𝑁−𝑍21 𝑍22 −𝑍2𝑁⋮
−𝑍𝑁1
⋮−𝑍𝑁2
⋱ ⋮⋯ 𝑍𝑁𝑁
𝐼1𝐼2⋮𝐼𝑁
=
𝑉1𝑉2⋮𝑉𝑁
𝑍𝑖𝑖 =𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑖𝑛 𝑙𝑜𝑜𝑝 𝑖
𝑍𝑖𝑗 =𝐶𝑜𝑚𝑚𝑜𝑛 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑙𝑜𝑜𝑝𝑠 𝑖 𝑎𝑛𝑑 𝑗 = 𝑍𝑗𝑖
𝑉𝑖 =𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑠𝑜𝑢𝑟𝑐𝑒𝑠 𝑖𝑛 𝑙𝑜𝑜𝑝 𝑖𝑉 is +ve if it supplies
current in the direction
of the loop current
Previously on ELCN102
Dr. Mohamed Refky 8
Node Analysis Method
The Node Analysis Method (Nodal Analysis) uses KCL to
generate a set of simultaneous equations.
1) Convert independent voltage sources into equivalent current
sources.
2) Identify the number of non simple nodes (𝑁) of the circuit.
3) Write an expression for the KCL at each 𝑁 − 1 Node
(exclude the ground node).
4) Solve the resultant simultaneous equations to get the voltages.
Previously on ELCN102
Dr. Mohamed Refky
Matrix Form
𝑌11 −𝑌12 ⋯ −𝑌1𝑁−𝑌21 𝑌22 −𝑌2𝑁⋮
−𝑌𝑁1
⋮−𝑌𝑁2
⋱ ⋮⋯ 𝑌𝑁𝑁
𝑉1𝑉2⋮𝑉𝑁
=
𝐼1𝐼2⋮𝐼𝑁
𝑌𝑖𝑖 =𝑎𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑛𝑜𝑑𝑒 𝑖
𝑌𝑖𝑗 =𝑐𝑜𝑚𝑚𝑜𝑛 𝑎𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑛𝑜𝑑𝑒 𝑖 𝑎𝑛𝑑 𝑗 = 𝑌𝑗𝑖
𝐼𝑖 =𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑠𝑜𝑢𝑟𝑐𝑒𝑠 𝑎𝑡 𝑛𝑜𝑑𝑒 𝑖 𝐼 is +ve if it supply
current into the node
Previously on ELCN102
Dr. Mohamed Refky
Superposition TheoremFor a linear circuit containing multiple independent sources, the
voltage across (or current through) any of its elements is the
algebraic sum of the voltages across (or currents through) that
element due to each independent source acting alone.
10∠30𝑜V 𝐼𝑎
5∠0𝑜A 𝐼𝑏
Total 𝐼 = 𝐼𝑎 + 𝐼𝑏
Previously on ELCN102
Dr. Mohamed Refky
Thevenin’s TheoremA linear two-terminal circuit, can be replaced by an equivalent
circuit consisting of a voltage source 𝑉𝑡ℎ in series with a
impedance 𝑍𝑡ℎ.
Previously on ELCN102
Dr. Mohamed Refky
Thevenin’s Theorem Steps1) Identify the load impedance and introduce two nodes 𝑎 and 𝑏
2) Remove the load impedance between node 𝑎 and 𝑏
3) Calculate the open circuit voltage between nodes 𝑎 and 𝑏.This voltage is 𝑉𝑡ℎ of the Thevenin equivalent circuit.
4) Set all the independent sources to zero (voltage sources are
SC and current sources are OC) and calculate the impedance
seen between nodes 𝑎 and 𝑏. This impedance is 𝑍𝑡ℎ of the
Thevenin equivalent circuit.
Previously on ELCN102
Dr. Mohamed Refky
Norton’s TheoremA linear two-terminal circuit can be replaced by equivalent circuit
consisting of a current source 𝐼𝑁 in parallel with a impedance 𝑍𝑁
Previously on ELCN102
Dr. Mohamed Refky
Norton’s Theorem Steps1) Identify the load impedance and introduce two nodes 𝑎 and 𝑏
2) Remove the load impedance between node 𝑎 and 𝑏 and set all
the independent sources to zero (voltage sources are SC and
current sources are OC) and calculate the impedance seen
between nodes 𝑎 and 𝑏. This resistance is 𝑍𝑁 of the Norton
equivalent circuit.
3) Replace the load impedance with a short circuit and calculate
the short circuit current between nodes 𝑎 and 𝑏. This current
is 𝐼𝑁 of the Norton equivalent circuit.
Previously on ELCN102
Dr. Mohamed Refky
Thevenin and Norton equivalent circuits
Thevenin equivalent circuit must be equivalent to Norton
equivalent circuit
𝑍𝑁 = 𝑍𝑡ℎ, 𝑉𝑡ℎ = 𝐼𝑁𝑍𝑁, 𝐼𝑁 =𝑉𝑡ℎ𝑍𝑡ℎ
→ 𝑍𝑡ℎ =𝑉𝑡ℎ𝐼𝑁
AC Power Analysis
Dr. Mohamed Refky
Instantaneous PowerInstantaneous power 𝑝 𝑡 is the product of the instantaneous
voltage 𝑣(𝑡) across the element and the instantaneous current 𝑖(𝑡)through it
𝑝 𝑡 = 𝑣 𝑡 × 𝑖 𝑡
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙𝑣 , 𝑖 𝑡 = 𝐼𝑚 cos 𝜔𝑡 + 𝜙𝑖
𝑝 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙𝑣 × 𝐼𝑚 cos 𝜔𝑡 + 𝜙𝑖
= 𝑉𝑚 × 𝐼𝑚 cos 𝜔𝑡 + 𝜙𝑣 cos 𝜔𝑡 + 𝜙
AC Power Analysis
Dr. Mohamed Refky
Instantaneous PowerInstantaneous power 𝑝 𝑡 is the product of the instantaneous
voltage 𝑣(𝑡) across the element and the instantaneous current 𝑖(𝑡)through it
𝑝 𝑡 = 𝑉𝑚 × 𝐼𝑚 cos 𝜔𝑡 + 𝜙𝑣 cos 𝜔𝑡 + 𝜙𝑖
𝑝 𝑡 =𝑉𝑚 × 𝐼𝑚
2cos 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖 + cos 𝜙𝑣 − 𝜙𝑖
Hint: cos𝐴 cos𝐵 =1
2cos 𝐴 + 𝐵 + cos 𝐴 − 𝐵
AC Power Analysis
Dr. Mohamed Refky
Instantaneous Power
𝑝 𝑡 =1
2𝑉𝑚𝐼𝑚 cos 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖 + cos 𝜙𝑣 − 𝜙𝑖
AC Power Analysis
Dr. Mohamed Refky
Average PowerAverage Power (𝑃𝑎𝑣) is the average of 𝑝 𝑡 over one period
𝑃𝑎𝑣 =1
𝑇න0
𝑇
𝑝 𝑡 𝑑𝑡
=1
𝑇න0
𝑇 𝑉𝑚𝐼𝑚2
cos 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖 + cos 𝜙𝑣 − 𝜙𝑖 𝑑𝑡
=𝑉𝑚𝐼𝑚2𝑇
sin 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖
2𝜔+ 𝑡 cos 𝜙𝑣 − 𝜙𝑖
0
𝑇
AC Power Analysis
Dr. Mohamed Refky
Average PowerAverage Power (𝑃𝑎𝑣) is the average of 𝑝 𝑡 over one period
𝑃𝑎𝑣 =𝑉𝑚𝐼𝑚2𝑇
sin 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖
2𝜔+ 𝑡 cos 𝜙𝑣 − 𝜙𝑖
0
𝑇
sin 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖
𝑡=0sin 𝜙𝑣 + 𝜙𝑖
sin 2𝜔𝑡 + 𝜙𝑣 + 𝜙𝑖𝑡=𝑇
sin 2𝜔𝑇 + 𝜙𝑣 + 𝜙𝑖
𝜔 =2𝜋
𝑇sin 2 2𝜋 + 𝜙𝑣 + 𝜙𝑖
sin 𝜙𝑣 + 𝜙𝑖
AC Power Analysis
Dr. Mohamed Refky
Average PowerAverage Power (𝑃𝑎𝑣) is the average of 𝑝 𝑡 over one period
𝑃𝑎𝑣 =𝑉𝑚𝐼𝑚2𝑇
𝑇 cos 𝜙𝑣 − 𝜙𝑖 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
AC Power Analysis
Dr. Mohamed Refky
Average PowerAverage Power (𝑃𝑎𝑣) is the average of 𝑝 𝑡 over one period
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
= 𝐴 cos 𝜙𝑣 − 𝜙𝑖
cos 𝜙𝑣 − 𝜙𝑖 is called the Power factor.
𝐴 =1
2𝑉𝑚𝐼𝑚 is referred to as the Apparent power.
Hint: cos 𝑥 is even function → cos 𝜙𝑣 − 𝜙𝑖 = cos 𝜙𝑖 − 𝜙𝑣
AC Power Analysis
Dr. Mohamed Refky
Average Power
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
For an impedance 𝑍
𝑉 = 𝑉𝑚∠𝜙𝑣, 𝐼 = 𝐼𝑚∠𝜙𝑖 → 𝑍 =𝑉
𝐼=𝑉𝑚𝐼𝑚
∠𝜙𝑣 − 𝜙𝑖 = 𝑍 ∠𝑍
𝑃𝑎𝑣 =1
2𝑍 × 𝐼𝑚 𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
=𝐼𝑚2
2𝑍 cos ∠𝑍
AC Power Analysis
Dr. Mohamed Refky
Average Power
𝑍 = 𝑍 ∠𝑍 = 𝑅 + 𝑗𝑋
𝑅 = 𝑍 cos ∠𝑍 , 𝑋 = 𝑍 sin ∠𝑍
𝑃𝑎𝑣 =𝐼𝑚2
2𝑍 cos ∠𝑍 =
𝐼𝑚2 𝑅
2
For impedance 𝑍, the average power is given by
𝑃𝑎𝑣 =𝐼𝑚2 𝑅
2
𝑅 = 𝑅𝑒 𝑍 is the resistive part of 𝑍
AC Power Analysis
Dr. Mohamed Refky
Average Power
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
For 𝜙𝑣 − 𝜙𝑖 = 0
𝑍 is pure resistive and the average power absorbed by the
impedance will be maximum.
For 𝜙𝑣 − 𝜙𝑖 = ±90𝑜
𝑍 is pure reactive and the average power absorbed by the
impedance will be zero.
AC Power Analysis
Dr. Mohamed Refky
Example (1)For a linear circuit, 𝑣 𝑡 = 120cos(377𝑡 + 45𝑜) , and 𝑖(𝑡)= 10cos(377𝑡 − 10𝑜). Calculate the average power.
AC Power Analysis
Dr. Mohamed Refky
Example (2)For the shown circuit, calculate the average power delivered by
the source and the average power absorbed by the circuit.
AC Power Analysis
Dr. Mohamed Refky
Example (3)Verify the power balance for the shown circuit
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareRoot Mean Square (𝑟𝑚𝑠) or Effective value of a varying signal is
defined as the value of DC signal that would produce the same
power dissipation in a resistive load
The 𝑒𝑓𝑓 value of a sinusoidal signal
DC Signal
𝑃𝑎𝑣 = 𝐼𝑒𝑓𝑓2 × 𝑅
Sinusoidal Signal
𝑃𝑎𝑣 =𝐼𝑚2
2× 𝑅
=
2
Maximum of the sinusoidal
𝐼𝑒𝑓𝑓2 =
𝐼𝑚2
2→ 𝐼𝑒𝑓𝑓 =
𝐼𝑚
2
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareRoot Mean Square (𝑟𝑚𝑠) or Effective value of a varying signal is
defined as the value of DC signal that would produce the same
power dissipation in a resistive load
DC Signal
𝑃𝑎𝑣 = 𝐼𝑒𝑓𝑓2 × 𝑅
Any periodic signal
𝑃𝑎𝑣 =1
𝑇න0
𝑇
𝑖 𝑡 𝑣 𝑡 𝑑𝑡
=𝑅
𝑇න0
𝑇
𝑖2 𝑡 𝑑𝑡𝐼𝑒𝑓𝑓 =1
𝑇න0
𝑇
𝑖2 𝑡 𝑑𝑡
AC Power Analysis
Dr. Mohamed Refky
Root Mean Square
The effective value is the (square) root of the mean (or average)
of the square of the periodic signal. Thus, the effective value is
often known as the root-mean-square value, or 𝒓𝒎𝒔 value for
short.
𝑋𝑒𝑓𝑓 = 𝑋𝑟𝑚𝑠
For any periodic function 𝑥 𝑡 , the 𝑒𝑓𝑓 value is given by
𝑋𝑒𝑓𝑓 =1
𝑇න0
𝑇
𝑥2 𝑡 𝑑𝑡
AC Power Analysis
Dr. Mohamed Refky
Root Mean Square
For a voltage signal 𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡
𝑣2 𝑡 = 𝑉𝑚2 cos2 𝜔𝑡
1
𝑇න0
𝑇
𝑣2 𝑡 𝑑𝑡 =𝑉𝑚2
𝑇න0
𝑇
cos2 𝜔𝑡 𝑑𝑡
For any periodic function 𝑥 𝑡 , the 𝑒𝑓𝑓 value is given by
𝑋𝑟𝑚𝑠 = 𝑋𝑒𝑓𝑓 =1
𝑇න0
𝑇
𝑥2 𝑡 𝑑𝑡
AC Power Analysis
Dr. Mohamed Refky
Root Mean Square
𝑉𝑚2
𝑇න0
𝑇
cos2 𝜔𝑡 𝑑𝑡 =𝑉𝑚2
2𝑇න0
𝑇
1 + cos 2𝜔𝑡 𝑑𝑡
=𝑉𝑚2
2𝑇𝑡 +
sin 2𝜔𝑡
2𝜔0
𝑇
=𝑉𝑚2
2
𝑉𝑟𝑚𝑠 =1
𝑇න0
𝑇
𝑣2 𝑡 𝑑𝑡 =𝑉𝑚
2
Hint:
cos2 𝜃 =1
21 + cos 2𝜃
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareFor a resistor the average power is given by
𝑃𝑎𝑣 =𝐼𝑚2
2𝑅 =
𝐼𝑚
2
𝐼𝑚
2𝑅
= 𝐼𝑟𝑚𝑠2 𝑅
=𝑉𝑟𝑚𝑠2
𝑅
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareFor a resistor the average power is given by
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖
= 𝐴 × 𝑝𝑓
𝐴 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 is called the Apparent power because it similar to
the average power of the DC circuit. Apparent power is measured
in volt-amperes (𝑉𝐴) to distinguish it from the average power,
which is measured in watts.
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareFor a resistor the average power is given by
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖
= 𝐴 × 𝑝𝑓
𝑝𝑓 = cos 𝜙𝑣 − 𝜙𝑖 is called the power factor. It is factor by
which the apparent power must be multiplied to obtain the
average power. For a purely resistive load, 𝑝𝑓 = 1. For a purely
reactive load, 𝑝𝑓 = 0.
AC Power Analysis
Dr. Mohamed Refky
Root Mean SquareFor a resistor the average power is given by
𝑃𝑎𝑣 =1
2𝑉𝑚𝐼𝑚 cos 𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖
= 𝐴 × 𝑝𝑓
𝑝𝑓 is said to be leading if the current leads the voltage, which
implies a capacitive load. 𝑝𝑓 is said to be lagging if the current
lags the voltage, which implies a inductive load.
AC Power Analysis
Dr. Mohamed Refky
Example (4)Determine the rms value of the current waveform shown. If the
current is passed through a 2Ω resistor, find the average power
absorbed by the resistor.
AC Power Analysis
Dr. Mohamed Refky
Complex PowerComplex power is important in power analysis because it
contains all the information about the power absorbed by a given
load.
The complex power 𝑆 absorbed by an AC load is the product of
the voltage and the complex conjugate of the current
𝑆 =1
2𝑉 × 𝐼∗
𝑉 = 𝑉𝑚∠𝜙𝑣, 𝐼 = 𝐼𝑚∠𝜙𝑖
𝑆 =1
2𝑉𝑚∠𝜙𝑣 𝐼𝑚∠ − 𝜙𝑖
AC Power Analysis
Dr. Mohamed Refky
Complex Power
𝑆 =1
2𝑉𝑚𝐼𝑚∠𝜙𝑣 − 𝜙𝑖 =
𝑉𝑚
2
𝐼𝑚
2∠𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
For an impedance 𝑍
𝑍 =𝑉
𝐼=𝑉𝑚∠𝜙𝑣𝐼𝑚∠𝜙𝑖
=𝑉𝑚∠𝜙𝑣 − 𝜙𝑖
𝐼𝑚=𝑉𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
𝐼𝑟𝑚𝑠
𝑆 = 𝑍 × 𝐼𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠2 𝑍 =
𝑉𝑟𝑚𝑠2
𝑍∗
= 𝐼𝑟𝑚𝑠2 𝑅 + 𝑗𝑋 = 𝐼𝑟𝑚𝑠
2 𝑅 + 𝑗𝐼𝑟𝑚𝑠2 𝑋
AC Power Analysis
Dr. Mohamed Refky
Complex Power
𝑆 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖 + 𝑗𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 sin 𝜙𝑣 − 𝜙𝑖
= 𝐼𝑟𝑚𝑠2 𝑅 + 𝑗𝐼𝑟𝑚𝑠
2 𝑋
= 𝑃 + 𝑗𝑄
𝑃 = 𝑅𝑒 𝑆 is called the Active Power and its unit is Watt.
𝑄 = 𝐼𝑚 𝑆 is called the Reactive Power and its unit is VAR.
𝐴 = 𝑆 = 𝑉𝑟𝑚𝑠 × 𝐼𝑟𝑚𝑠 is the Apparent Power and its unit is VA.
𝜙𝑣 − 𝜙𝑖 is the Power Factor Angle.
AC Power Analysis
Dr. Mohamed Refky
Complex Power
𝑆 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖 + 𝑗𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 sin 𝜙𝑣 − 𝜙𝑖
= 𝐼𝑟𝑚𝑠2 𝑅 + 𝑗𝐼𝑟𝑚𝑠
2 𝑋
= 𝑃 + 𝑗𝑄
The real power 𝑃 is the average power in watts delivered to a
load. 𝑃 is the actual power dissipated by the load.
The reactive power 𝑄 is a measure of the energy exchange
between the source and the reactive part of the load.
AC Power Analysis
Dr. Mohamed Refky
Complex Power
𝑆 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖 + 𝑗𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 sin 𝜙𝑣 − 𝜙𝑖
= 𝐼𝑟𝑚𝑠2 𝑅 + 𝑗𝐼𝑟𝑚𝑠
2 𝑋
= 𝑃 + 𝑗𝑄
For resistive load, 𝜙𝑣 − 𝜙𝑖 = 0 → 𝑄 = 0.
For capacitive load (leading pf), 𝜙𝑣 − 𝜙𝑖 < 0 → 𝑄 < 0.
For inductive load (lagging pf), 𝜙𝑣 − 𝜙𝑖 > 0 → 𝑄 > 0.
AC Power Analysis
Dr. Mohamed Refky
Complex Power
𝑆 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠∠𝜙𝑣 − 𝜙𝑖
= 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜙𝑣 − 𝜙𝑖 + 𝑗𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 sin 𝜙𝑣 − 𝜙𝑖
= 𝐼𝑟𝑚𝑠2 𝑅 + 𝑗𝐼𝑟𝑚𝑠
2 𝑋
= 𝑃 + 𝑗𝑄
𝑆, 𝑃, and 𝑄 construct the
power triangle.
AC Power Analysis
Dr. Mohamed Refky
Example (5)The voltage across a load is 𝑣(𝑡) = 60cos(𝜔𝑡 − 10𝑜) V and the
current through the element in the direction of the voltage drop is
𝑖(𝑡) = 1.5cos(𝜔𝑡 + 50) A. Find:
(a) The complex and apparent powers.
(b) The real and reactive powers.
(c) The power factor and the load impedance.
Maximum Average Power Transfer
Dr. Mohamed Refky
For maximum average power transfer, the load impedance 𝑍𝐿must be equal to the complex conjugate of the Thevenin/Norton
impedance 𝑍𝑇ℎ.
→ 𝑍𝐿 = 𝑍𝑡ℎ∗ = 𝑍𝑁
∗For maximum power 𝑃𝑍𝐿
Definition
Maximum Average Power Transfer
Dr. Mohamed Refky
𝑍𝑡ℎ = 𝑅𝑡ℎ + 𝑗𝑋𝑡ℎ
𝑍𝐿 = 𝑍𝑡ℎ∗ = 𝑅𝑡ℎ − 𝑗𝑋𝑡ℎ
𝑍𝑇 = 𝑍𝑡ℎ + 𝑍𝐿 = 2𝑅𝑡ℎ
𝐼 =𝑉𝑡ℎ𝑍𝑇
=𝑉𝑡ℎ2𝑅𝑡ℎ
𝑉𝑡ℎ = 𝑉𝑚∠0𝑜
𝐼 =𝑉𝑚2𝑅𝑡ℎ
∠0𝑜
Definition
𝑃𝑍𝐿 =𝐼𝑚2
2𝑅𝑡ℎ
=𝑉𝑚2𝑅𝑡ℎ
2𝑅𝑡ℎ2
Maximum Average Power Transfer
Dr. Mohamed Refky
𝑃𝑍𝐿 =𝑉𝑚2
4𝑅𝑡ℎ2
𝑅𝑡ℎ2
𝑃𝑍𝐿 =𝑉𝑚2
8𝑅𝑡ℎ
𝑉𝑟𝑚𝑠 =𝑉𝑚
2→ 𝑉𝑟𝑚𝑠
2 =𝑉𝑚2
2
𝑃𝑍𝐿 =𝑉𝑟𝑚𝑠2
4𝑅𝑡ℎ
Definition
𝑃𝑍𝐿 =𝐼𝑚2
2𝑅𝑡ℎ
=𝑉𝑚2𝑅𝑡ℎ
2𝑅𝑡ℎ2
Maximum Average Power Transfer
Dr. Mohamed Refky
Example (6)For the circuit shown:
a) Determine the load impedance 𝑍𝐿 that maximizes the average
power drawn from the circuit.
b) What is the maximum average power ?