PY2P10 Current Electricity - Nicky
Transcript of PY2P10 Current Electricity - Nicky
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Your course, please
A. Science
B. Nanoscience
C. Theoretical Physics
Session ID: PY2P10EM Laptop: responseware.eu
AC Sources
β’ Most present-day household and industrial power distribution systems operate with alternating current (ac).
β’ Any appliance that you plug into a wall outlet uses ac.
β’ 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 π‘ with period π =2π
π:
π£ π‘ + ππ = ππ cos π π‘ + ππ + π = ππ cos ππ‘ + π + πππ
= ππ cos ππ‘ + π + ππ2π
π= ππcos ππ‘ + π + π2π
= ππcos(ππ‘ + π) = π£(π‘)
π£ π‘ = ππ cos ππ‘ + πSinusoidal
the amplitude
angular frequency measured in radians/s and the cyclic frequency π in Hz, π = 2ππ
the phase
Two sinusoids cos(π₯ + π/3) leads cos(π₯) by π/3
cos(π₯ β π/3) lags cos π₯ by π/3
π£1 = cos π₯ +π
3, π1 =
π
3
π£2 = cos π₯, π2 = 0
π£1 leads π£2 by
Ξπ = π1 β π2 =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.
β’ Ξπ > 0, π£1 leads π£2 by Ξπβ’ Ξπ = 0, π£1 and π£2 are in phase
β’ Ξπ < 0, π£1 lags π£2 by Ξπ
Ξπ = π1 β π2
Phase angle
between two
signals:
π£1 = ππ1 cos(ππ‘ + π1)
π£2 = ππ2 cos(ππ‘ + π2)
Positive
Use cosine
βcos π₯ = cos(π₯ Β± π)
sin(π₯) = cos(π₯ βπ
2)
βπ < π < π
Q1: The phase angle between,
π£1 = β10 cos(ππ‘ +π
3) and
π£2 = 12 sin ππ‘ +π
6
is: π
3β
π
6=
π
6. This means π£1 leads π£2 by
π
6.
A. True
B. False
Answer to Q1:
Use the same form:
π£1 = β10 cos ππ‘ +π
3= 10 cos ππ‘ +
π
3β π = 10 cos ππ‘ β
2π
3
π£2 = 12 sin ππ‘ +π
6= 12 cos(ππ‘ +
π
6βπ
2) = 12 cos(ππ‘ β
π
3)
Compare:
Ξπ = β2π
3βπ
3= β
π
3< 0
Therefore, π£1 lags π£2 by π
3.
Phasor
β’ A phasor is a complex number that represents the amplitude (e.g. ππ) and phase (π) of a sinusoid, π£ π‘ = ππ cos(ππ‘ + π).
β’ The real part of a phasor represents the sinusoid signal π£ π‘ .
β’ Since we consider a single frequency, the phasor can be written as π½ = πππ
ππ , i.e. ππππ‘ is implicitly present.
Phasor : ππππ ππ‘+π = ππ cos(ππ‘ + π) + πππ sin(ππ‘ + π)
Exponential representation Rectangular representation
The sinusoid signal π£(π‘)
Q2: The phasor of π£1 = β10 cos(ππ‘ +π
3) is
A. β10πππ
3
B. 10πππ
3
C. 10πβππ
3
D. 10πβπ2π
3
Q3: The ac sinusoid voltage π£(π‘) (π = 4 rads/s) that corresponds to a phasor π½ = 3V is
A. 3 V
B. cos(4π‘ + 3)V
C. 3 cos(4π‘)V
D. 3 sin(4π‘) V
E. 3 cos(4π‘ + π)V
Answer to Q2-Q3
Q2: To wirte the phasor for π£1 = β10 cos(ππ‘ +π
3) , we first need to
convert it into the conventional form, i.e. a cosine with a positive amplitude,
π£1 = β10 cos ππ‘ +π
3= 10 cos ππ‘ +
π
3β π = 10 cos(ππ‘ β
2π
3)
Therefore, π½ = 10πβ2π
3π = 10β β 120O
Q3: Note π½ = 3 = 3eπβ 0, i.e. the amplitude is 3 V, the phase is 0, and π = 4
Thereforeπ£ π‘ = 3 cos ππ‘ + 0 = 3 cos(4π‘) V
Phasor diagram
β’ To represent sinusoidallyvarying voltages and currents, we define rotating vectors in the Argand plane called phasors.
β’ Shown is a phasor diagramfor sinusoidal voltage and current with their initial phases π and βπ.
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 π, i.e. we consider signals which have the same frequency. Circuit response depends on π. If we switch from one frequency to another, the circuit responses changes.
πππππ
Time-independent and complex
Phasor Time
Time-dependent and real
ππ cos(ππ‘ + π)
Derivative and integral in phasor domain
In phasor representation, the time derivative of a sinusoid becomes just multiplication by the constant ππ; integrating a
phasor corresponds to multiplication by 1
ππ.
PhasorTime
π£ π‘ = ππ cos(ππ‘ + π)ππ£ π‘
ππ‘= βπππ sin ππ‘ + π = πππ cos(ππ‘ + π +
π
2)
ΰΆ±π£ π‘ ππ‘ =πππsin(ππ‘ + π) =
πππcos(ππ‘ + π β
π
2)
π½ = πππππ
πππππ π+
π2 = (πππ
ππ)ππππ2 = πππ½
πππππ πβ
π2 = πππ
πππβπ
π2
π=
π½
ππ
Try this: π ππππ ππ‘+π
ππ‘
Q4: In an ac circuit, the voltage across a 4Ξ©resistor is π£ π‘ = 4 cos(10π‘ + π/3), the phase of the current through the resistor is
A. 0
B. βπ
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
β’ The resistance does not depend on the frequency of the ac source.
β’ The voltage and current are related by Ohmβs law: π£π (π‘) = ππ π‘ π and Ohmβs law holds in phasor domain.
β’ Current and voltage are in phase.
PhasorTime
ππ π‘ = πΌπ cos(ππ‘ + π)π£π π‘ = ππ π‘ π = πΌππ cos(ππ‘ + π)
π°πΉ = πΌππππ
π½πΉ = πΌππ πππ = π π°πΉ
Q6: For an inductor, its voltage and current are always in phase.
A. True
B. False
Inductor in an ac circuit
β’ The inductance does not depend on the frequency of the ac source.
β’ The voltage and current are related by :
π£πΏ(π‘) = πΏπππΏ π‘
ππ‘.
β’ Voltage leads current by π/2
PhasorTime
ππΏ π‘ = πΌπ cos(ππ‘ + π)
π£πΏ π‘ = πΏπππΏ π‘
ππ‘= βπΏππΌπ sin(ππ‘ + π)
π°π³ = πΌππππ
π½π³ = (πππΏ)π°π³
Q7: For a capacitor, its voltage and current are always in phase.
A. True
B. False
Capacitor in an ac circuit
β’ The capacitance does not depend on the frequency of the ac source.
β’ The voltage and current are related by :
ππΆ(π‘) = πΆππ£πΆ π‘
ππ‘.
β’ Current leads voltage by π/2.
PhasorTime
π£π π‘ = ππ cos(ππ‘ + π)
ππΆ π‘ = πΆππ£π π‘
ππ‘= βπΆπππ sin(ππ‘ + π)
π°πͺ = πΌππππ
π½πͺ =π°πͺπππΆ
Impedance and admittance
β’ Impedance represents the opposition to the flow of sinusoidal current.
β’ π is generally a complex number (Ξ©).
β’ Admittance, π = π/π is the inverse of impedance (S).
π½ = ππ°
π½ = π π°π½ = πππΏ π°
π½ =1
πππΆπ°
Phasor
Impedance
Continued on next page
β’ π < 0, capacitive/leading reactance, e.g. π = β1
ππΆπ
β’ π > 0, inductive/lagging reactance, e.g. π = ππΏ π
Impedance:
Admittance:
π = π + ππ
π =π
π= πΊ + ππ΅
Resistance Reactance
Conductance Susceptance
Capacitor Inductor
π β 0 π = β1
ππΆπ β β
open circuit
π = ππΏ π β 0short circuit
π β β π = β1
ππΆπ β 0
short circuit
π = ππΏ π β βopen circuit
Circuit response depends on the frequency