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Your course, please
A. Science
B. Nanoscience
C. Theoretical Physics
Session ID: PY2P10EM Laptop: responseware.eu
AC Sources
• Most present-day household and industrial power distribution systems operate with alternating current (ac).
• Any appliance that you plug into a wall outlet uses ac.
• An ac source is a device that supplies a sinusoidally varying voltage.
Sinusoids
A sinusoidal is a signal that has the form of the sine or cosine function, which is a time-varying excitation
It is a period function of time 𝑡 with period 𝑇 =2𝜋
𝜔:
𝑣 𝑡 + 𝑛𝑇 = 𝑉𝑚 cos 𝜔 𝑡 + 𝑛𝑇 + 𝜙 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙 + 𝑛𝜔𝑇
= 𝑉𝑚 cos 𝜔𝑡 + 𝜙 + 𝑛𝜔2𝜋
𝜔= 𝑉𝑚cos 𝜔𝑡 + 𝜙 + 𝑛2𝜋
= 𝑉𝑚cos(𝜔𝑡 + 𝜙) = 𝑣(𝑡)
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙Sinusoidal
the amplitude
angular frequency measured in radians/s and the cyclic frequency 𝑓 in Hz, 𝜔 = 2𝜋𝑓
the phase
Two sinusoids cos(𝑥 + 𝜋/3) leads cos(𝑥) by 𝜋/3
cos(𝑥 − 𝜋/3) lags cos 𝑥 by 𝜋/3
𝑣1 = cos 𝑥 +𝜋
3, 𝜙1 =
𝜋
3
𝑣2 = cos 𝑥, 𝜙2 = 0
𝑣1 leads 𝑣2 by
Δ𝜙 = 𝜙1 − 𝜙2 =𝜋
3
Phase angleA sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.
• Δ𝜙 > 0, 𝑣1 leads 𝑣2 by Δ𝜙• Δ𝜙 = 0, 𝑣1 and 𝑣2 are in phase
• Δ𝜙 < 0, 𝑣1 lags 𝑣2 by Δ𝜙
Δ𝜙 = 𝜙1 − 𝜙2
Phase angle
between two
signals:
𝑣1 = 𝑉𝑚1 cos(𝜔𝑡 + 𝜙1)
𝑣2 = 𝑉𝑚2 cos(𝜔𝑡 + 𝜙2)
Positive
Use cosine
−cos 𝑥 = cos(𝑥 ± 𝜋)
sin(𝑥) = cos(𝑥 −𝜋
2)
−𝝅 < 𝝓 < 𝝅
Q1: The phase angle between,
𝑣1 = −10 cos(𝜔𝑡 +𝜋
3) and
𝑣2 = 12 sin 𝜔𝑡 +𝜋
6
is: 𝜋
3−
𝜋
6=
𝜋
6. This means 𝑣1 leads 𝑣2 by
𝜋
6.
A. True
B. False
Answer to Q1:
Use the same form:
𝑣1 = −10 cos 𝜔𝑡 +𝜋
3= 10 cos 𝜔𝑡 +
𝜋
3− 𝜋 = 10 cos 𝜔𝑡 −
2𝜋
3
𝑣2 = 12 sin 𝜔𝑡 +𝜋
6= 12 cos(𝜔𝑡 +
𝜋
6−𝜋
2) = 12 cos(𝜔𝑡 −
𝜋
3)
Compare:
Δ𝜙 = −2𝜋
3−𝜋
3= −
𝜋
3< 0
Therefore, 𝑣1 lags 𝑣2 by 𝜋
3.
Phasor
• A phasor is a complex number that represents the amplitude (e.g. 𝑉𝑚) and phase (𝜙) of a sinusoid, 𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙).
• The real part of a phasor represents the sinusoid signal 𝑣 𝑡 .
• Since we consider a single frequency, the phasor can be written as 𝑽 = 𝑉𝑚𝑒
𝑖𝜙 , i.e. 𝑒𝑖𝜔𝑡 is implicitly present.
Phasor : 𝑉𝑚𝑒𝑖 𝜔𝑡+𝜙 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙) + 𝑖𝑉𝑚 sin(𝜔𝑡 + 𝜙)
Exponential representation Rectangular representation
The sinusoid signal 𝑣(𝑡)
Q2: The phasor of 𝑣1 = −10 cos(𝜔𝑡 +𝜋
3) is
A. −10𝑒𝑖𝜋
3
B. 10𝑒𝑖𝜋
3
C. 10𝑒−𝑖𝜋
3
D. 10𝑒−𝑖2𝜋
3
Q3: The ac sinusoid voltage 𝑣(𝑡) (𝜔 = 4 rads/s) that corresponds to a phasor 𝑽 = 3V is
A. 3 V
B. cos(4𝑡 + 3)V
C. 3 cos(4𝑡)V
D. 3 sin(4𝑡) V
E. 3 cos(4𝑡 + 𝜋)V
Answer to Q2-Q3
Q2: To wirte the phasor for 𝑣1 = −10 cos(𝜔𝑡 +𝜋
3) , we first need to
convert it into the conventional form, i.e. a cosine with a positive amplitude,
𝑣1 = −10 cos 𝜔𝑡 +𝜋
3= 10 cos 𝜔𝑡 +
𝜋
3− 𝜋 = 10 cos(𝜔𝑡 −
2𝜋
3)
Therefore, 𝑽 = 10𝑒−2𝜋
3𝑖 = 10∠ − 120O
Q3: Note 𝑽 = 3 = 3e𝑖⋅0, i.e. the amplitude is 3 V, the phase is 0, and 𝜔 = 4
Therefore𝑣 𝑡 = 3 cos 𝜔𝑡 + 0 = 3 cos(4𝑡) V
Phasor diagram
• To represent sinusoidallyvarying voltages and currents, we define rotating vectors in the Argand plane called phasors.
• Shown is a phasor diagramfor sinusoidal voltage and current with their initial phases 𝜙 and −𝜃.
Time domain and phasor (frequency) domain
Time-domain representation is time dependent and always real, and its phasor (or frequency) domain counterpart is time-independent, generally complex. The phasor domain is for a constant 𝜔, i.e. we consider signals which have the same frequency. Circuit response depends on 𝜔. If we switch from one frequency to another, the circuit responses changes.
𝑉𝑚𝑒𝑖𝜙
Time-independent and complex
Phasor Time
Time-dependent and real
𝑉𝑚 cos(𝜔𝑡 + 𝜙)
Derivative and integral in phasor domain
In phasor representation, the time derivative of a sinusoid becomes just multiplication by the constant 𝑖𝜔; integrating a
phasor corresponds to multiplication by 1
𝑖𝜔.
PhasorTime
𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)𝑑𝑣 𝑡
𝑑𝑡= −𝑉𝑚𝜔 sin 𝜔𝑡 + 𝜙 = 𝑉𝑚𝜔 cos(𝜔𝑡 + 𝜙 +
𝜋
2)
න𝑣 𝑡 𝑑𝑡 =𝑉𝑚𝜔sin(𝜔𝑡 + 𝜙) =
𝑉𝑚𝜔cos(𝜔𝑡 + 𝜙 −
𝜋
2)
𝑽 = 𝑉𝑚𝑒𝑖𝜙
𝑉𝑚𝜔𝑒𝑖 𝜙+
𝜋2 = (𝑉𝑚𝑒
𝑖𝜙)𝜔𝑒𝑖𝜋2 = 𝑖𝜔𝑽
𝑉𝑚𝜔𝑒𝑖 𝜙−
𝜋2 = 𝑉𝑚𝑒
𝑖𝜙𝑒−𝑖
𝜋2
𝜔=
𝑽
𝑖𝜔
Try this: 𝑑 𝑉𝑚𝑒𝑖 𝜔𝑡+𝜙
𝑑𝑡
Q4: In an ac circuit, the voltage across a 4Ωresistor is 𝑣 𝑡 = 4 cos(10𝑡 + 𝜋/3), the phase of the current through the resistor is
A. 0
B. −𝜋
3
C.𝜋
3
D. None of the above
Q5: For a resistor, its voltage and current are always in phase.
A. True
B. False
Resistor in an ac circuit
• The resistance does not depend on the frequency of the ac source.
• The voltage and current are related by Ohm’s law: 𝑣𝑅(𝑡) = 𝑖𝑅 𝑡 𝑅 and Ohm’s law holds in phasor domain.
• Current and voltage are in phase.
PhasorTime
𝑖𝑅 𝑡 = 𝐼𝑚 cos(𝜔𝑡 + 𝜙)𝑣𝑅 𝑡 = 𝑖𝑅 𝑡 𝑅 = 𝐼𝑚𝑅 cos(𝜔𝑡 + 𝜙)
𝑰𝑹 = 𝐼𝑚𝑒𝑖𝜙
𝑽𝑹 = 𝐼𝑚𝑅𝑒𝑖𝜙 = 𝑅𝑰𝑹
Q6: For an inductor, its voltage and current are always in phase.
A. True
B. False
Inductor in an ac circuit
• The inductance does not depend on the frequency of the ac source.
• The voltage and current are related by :
𝑣𝐿(𝑡) = 𝐿𝑑𝑖𝐿 𝑡
𝑑𝑡.
• Voltage leads current by 𝜋/2
PhasorTime
𝑖𝐿 𝑡 = 𝐼𝑚 cos(𝜔𝑡 + 𝜙)
𝑣𝐿 𝑡 = 𝐿𝑑𝑖𝐿 𝑡
𝑑𝑡= −𝐿𝜔𝐼𝑚 sin(𝜔𝑡 + 𝜙)
𝑰𝑳 = 𝐼𝑚𝑒𝑖𝜙
𝑽𝑳 = (𝑖𝜔𝐿)𝑰𝑳
Q7: For a capacitor, its voltage and current are always in phase.
A. True
B. False
Capacitor in an ac circuit
• The capacitance does not depend on the frequency of the ac source.
• The voltage and current are related by :
𝑖𝐶(𝑡) = 𝐶𝑑𝑣𝐶 𝑡
𝑑𝑡.
• Current leads voltage by 𝜋/2.
PhasorTime
𝑣𝑐 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)
𝑖𝐶 𝑡 = 𝐶𝑑𝑣𝑐 𝑡
𝑑𝑡= −𝐶𝜔𝑉𝑚 sin(𝜔𝑡 + 𝜙)
𝑰𝑪 = 𝐼𝑚𝑒𝑖𝜙
𝑽𝑪 =𝑰𝑪𝑖𝜔𝐶
Impedance and admittance
• Impedance represents the opposition to the flow of sinusoidal current.
• 𝒁 is generally a complex number (Ω).
• Admittance, 𝒀 = 𝟏/𝒁 is the inverse of impedance (S).
𝑽 = 𝒁𝑰
𝑽 = 𝑅𝑰𝑽 = 𝑖𝜔𝐿 𝑰
𝑽 =1
𝑖𝜔𝐶𝑰
Phasor
Impedance
Continued on next page
• 𝑋 < 0, capacitive/leading reactance, e.g. 𝒁 = −1
𝜔𝐶𝑖
• 𝑋 > 0, inductive/lagging reactance, e.g. 𝒁 = 𝜔𝐿 𝑖
Impedance:
Admittance:
𝒁 = 𝑅 + 𝑖𝑋
𝒀 =𝟏
𝒁= 𝐺 + 𝑖𝐵
Resistance Reactance
Conductance Susceptance
Capacitor Inductor
𝜔 → 0 𝒁 = −1
𝜔𝐶𝑖 → ∞
open circuit
𝒁 = 𝜔𝐿 𝑖 → 0short circuit
𝜔 → ∞ 𝒁 = −1
𝜔𝐶𝑖 → 0
short circuit
𝒁 = 𝜔𝐿 𝑖 → ∞open circuit
Circuit response depends on the frequency