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 𝑑 with period 𝑇 =2πœ‹

πœ”:

𝑣 𝑑 + 𝑛𝑇 = π‘‰π‘š cos πœ” 𝑑 + 𝑛𝑇 + πœ™ = π‘‰π‘š cos πœ”π‘‘ + πœ™ + π‘›πœ”π‘‡

= π‘‰π‘š cos πœ”π‘‘ + πœ™ + π‘›πœ”2πœ‹

πœ”= π‘‰π‘šcos πœ”π‘‘ + πœ™ + 𝑛2πœ‹

= π‘‰π‘šcos(πœ”π‘‘ + πœ™) = 𝑣(𝑑)

𝑣 𝑑 = π‘‰π‘š cos πœ”π‘‘ + πœ™Sinusoidal

the amplitude

angular frequency measured in radians/s and the cyclic frequency 𝑓 in Hz, πœ” = 2πœ‹π‘“

the phase

Two sinusoids cos(π‘₯ + πœ‹/3) leads cos(π‘₯) by πœ‹/3

cos(π‘₯ βˆ’ πœ‹/3) lags cos π‘₯ by πœ‹/3

𝑣1 = cos π‘₯ +πœ‹

3, πœ™1 =

πœ‹

3

𝑣2 = cos π‘₯, πœ™2 = 0

𝑣1 leads 𝑣2 by

Ξ”πœ™ = πœ™1 βˆ’ πœ™2 =πœ‹

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€’ Ξ”πœ™ > 0, 𝑣1 leads 𝑣2 by Ξ”πœ™β€’ Ξ”πœ™ = 0, 𝑣1 and 𝑣2 are in phase

β€’ Ξ”πœ™ < 0, 𝑣1 lags 𝑣2 by Ξ”πœ™

Ξ”πœ™ = πœ™1 βˆ’ πœ™2

Phase angle

between two

signals:

𝑣1 = π‘‰π‘š1 cos(πœ”π‘‘ + πœ™1)

𝑣2 = π‘‰π‘š2 cos(πœ”π‘‘ + πœ™2)

Positive

Use cosine

βˆ’cos π‘₯ = cos(π‘₯ Β± πœ‹)

sin(π‘₯) = cos(π‘₯ βˆ’πœ‹

2)

βˆ’π… < 𝝓 < 𝝅

Q1: The phase angle between,

𝑣1 = βˆ’10 cos(πœ”π‘‘ +πœ‹

3) and

𝑣2 = 12 sin πœ”π‘‘ +πœ‹

6

is: πœ‹

3βˆ’

πœ‹

6=

πœ‹

6. This means 𝑣1 leads 𝑣2 by

πœ‹

6.

A. True

B. False

Answer to Q1:

Use the same form:

𝑣1 = βˆ’10 cos πœ”π‘‘ +πœ‹

3= 10 cos πœ”π‘‘ +

πœ‹

3βˆ’ πœ‹ = 10 cos πœ”π‘‘ βˆ’

2πœ‹

3

𝑣2 = 12 sin πœ”π‘‘ +πœ‹

6= 12 cos(πœ”π‘‘ +

πœ‹

6βˆ’πœ‹

2) = 12 cos(πœ”π‘‘ βˆ’

πœ‹

3)

Compare:

Ξ”πœ™ = βˆ’2πœ‹

3βˆ’πœ‹

3= βˆ’

πœ‹

3< 0

Therefore, 𝑣1 lags 𝑣2 by πœ‹

3.

Phasor

β€’ A phasor is a complex number that represents the amplitude (e.g. π‘‰π‘š) and phase (πœ™) of a sinusoid, 𝑣 𝑑 = π‘‰π‘š cos(πœ”π‘‘ + πœ™).

β€’ The real part of a phasor represents the sinusoid signal 𝑣 𝑑 .

β€’ Since we consider a single frequency, the phasor can be written as 𝑽 = π‘‰π‘šπ‘’

π‘–πœ™ , i.e. π‘’π‘–πœ”π‘‘ is implicitly present.

Phasor : π‘‰π‘šπ‘’π‘– πœ”π‘‘+πœ™ = π‘‰π‘š cos(πœ”π‘‘ + πœ™) + π‘–π‘‰π‘š sin(πœ”π‘‘ + πœ™)

Exponential representation Rectangular representation

The sinusoid signal 𝑣(𝑑)

Q2: The phasor of 𝑣1 = βˆ’10 cos(πœ”π‘‘ +πœ‹

3) is

A. βˆ’10π‘’π‘–πœ‹

3

B. 10π‘’π‘–πœ‹

3

C. 10π‘’βˆ’π‘–πœ‹

3

D. 10π‘’βˆ’π‘–2πœ‹

3

Q3: The ac sinusoid voltage 𝑣(𝑑) (πœ” = 4 rads/s) that corresponds to a phasor 𝑽 = 3V is

A. 3 V

B. cos(4𝑑 + 3)V

C. 3 cos(4𝑑)V

D. 3 sin(4𝑑) V

E. 3 cos(4𝑑 + πœ‹)V

Answer to Q2-Q3

Q2: To wirte the phasor for 𝑣1 = βˆ’10 cos(πœ”π‘‘ +πœ‹

3) , we first need to

convert it into the conventional form, i.e. a cosine with a positive amplitude,

𝑣1 = βˆ’10 cos πœ”π‘‘ +πœ‹

3= 10 cos πœ”π‘‘ +

πœ‹

3βˆ’ πœ‹ = 10 cos(πœ”π‘‘ βˆ’

2πœ‹

3)

Therefore, 𝑽 = 10π‘’βˆ’2πœ‹

3𝑖 = 10∠ βˆ’ 120O

Q3: Note 𝑽 = 3 = 3e𝑖⋅0, i.e. the amplitude is 3 V, the phase is 0, and πœ” = 4

Therefore𝑣 𝑑 = 3 cos πœ”π‘‘ + 0 = 3 cos(4𝑑) 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œ™ and βˆ’πœƒ.

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œ”, i.e. we consider signals which have the same frequency. Circuit response depends on πœ”. If we switch from one frequency to another, the circuit responses changes.

π‘‰π‘šπ‘’π‘–πœ™

Time-independent and complex

Phasor Time

Time-dependent and real

π‘‰π‘š cos(πœ”π‘‘ + πœ™)

Derivative and integral in phasor domain

In phasor representation, the time derivative of a sinusoid becomes just multiplication by the constant π‘–πœ”; integrating a

phasor corresponds to multiplication by 1

π‘–πœ”.

PhasorTime

𝑣 𝑑 = π‘‰π‘š cos(πœ”π‘‘ + πœ™)𝑑𝑣 𝑑

𝑑𝑑= βˆ’π‘‰π‘šπœ” sin πœ”π‘‘ + πœ™ = π‘‰π‘šπœ” cos(πœ”π‘‘ + πœ™ +

πœ‹

2)

ࢱ𝑣 𝑑 𝑑𝑑 =π‘‰π‘šπœ”sin(πœ”π‘‘ + πœ™) =

π‘‰π‘šπœ”cos(πœ”π‘‘ + πœ™ βˆ’

πœ‹

2)

𝑽 = π‘‰π‘šπ‘’π‘–πœ™

π‘‰π‘šπœ”π‘’π‘– πœ™+

πœ‹2 = (π‘‰π‘šπ‘’

π‘–πœ™)πœ”π‘’π‘–πœ‹2 = π‘–πœ”π‘½

π‘‰π‘šπœ”π‘’π‘– πœ™βˆ’

πœ‹2 = π‘‰π‘šπ‘’

π‘–πœ™π‘’βˆ’π‘–

πœ‹2

πœ”=

𝑽

π‘–πœ”

Try this: 𝑑 π‘‰π‘šπ‘’π‘– πœ”π‘‘+πœ™

𝑑𝑑

Q4: In an ac circuit, the voltage across a 4Ξ©resistor is 𝑣 𝑑 = 4 cos(10𝑑 + πœ‹/3), the phase of the current through the resistor is

A. 0

B. βˆ’πœ‹

3

C.πœ‹

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 Ohm’s law: 𝑣𝑅(𝑑) = 𝑖𝑅 𝑑 𝑅 and Ohm’s law holds in phasor domain.

β€’ Current and voltage are in phase.

PhasorTime

𝑖𝑅 𝑑 = πΌπ‘š cos(πœ”π‘‘ + πœ™)𝑣𝑅 𝑑 = 𝑖𝑅 𝑑 𝑅 = πΌπ‘šπ‘… cos(πœ”π‘‘ + πœ™)

𝑰𝑹 = πΌπ‘šπ‘’π‘–πœ™

𝑽𝑹 = πΌπ‘šπ‘…π‘’π‘–πœ™ = 𝑅𝑰𝑹

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 :

𝑣𝐿(𝑑) = 𝐿𝑑𝑖𝐿 𝑑

𝑑𝑑.

β€’ Voltage leads current by πœ‹/2

PhasorTime

𝑖𝐿 𝑑 = πΌπ‘š cos(πœ”π‘‘ + πœ™)

𝑣𝐿 𝑑 = 𝐿𝑑𝑖𝐿 𝑑

𝑑𝑑= βˆ’πΏπœ”πΌπ‘š sin(πœ”π‘‘ + πœ™)

𝑰𝑳 = πΌπ‘šπ‘’π‘–πœ™

𝑽𝑳 = (π‘–πœ”πΏ)𝑰𝑳

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 :

𝑖𝐢(𝑑) = 𝐢𝑑𝑣𝐢 𝑑

𝑑𝑑.

β€’ Current leads voltage by πœ‹/2.

PhasorTime

𝑣𝑐 𝑑 = π‘‰π‘š cos(πœ”π‘‘ + πœ™)

𝑖𝐢 𝑑 = 𝐢𝑑𝑣𝑐 𝑑

𝑑𝑑= βˆ’πΆπœ”π‘‰π‘š sin(πœ”π‘‘ + πœ™)

𝑰π‘ͺ = πΌπ‘šπ‘’π‘–πœ™

𝑽π‘ͺ =𝑰π‘ͺπ‘–πœ”πΆ

Impedance and admittance

β€’ Impedance represents the opposition to the flow of sinusoidal current.

β€’ 𝒁 is generally a complex number (Ξ©).

β€’ Admittance, 𝒀 = 𝟏/𝒁 is the inverse of impedance (S).

𝑽 = 𝒁𝑰

𝑽 = 𝑅𝑰𝑽 = π‘–πœ”πΏ 𝑰

𝑽 =1

π‘–πœ”πΆπ‘°

Phasor

Impedance

Continued on next page

β€’ 𝑋 < 0, capacitive/leading reactance, e.g. 𝒁 = βˆ’1

πœ”πΆπ‘–

β€’ 𝑋 > 0, inductive/lagging reactance, e.g. 𝒁 = πœ”πΏ 𝑖

Impedance:

Admittance:

𝒁 = 𝑅 + 𝑖𝑋

𝒀 =𝟏

𝒁= 𝐺 + 𝑖𝐡

Resistance Reactance

Conductance Susceptance

Capacitor Inductor

πœ” β†’ 0 𝒁 = βˆ’1

πœ”πΆπ‘– β†’ ∞

open circuit

𝒁 = πœ”πΏ 𝑖 β†’ 0short circuit

πœ” β†’ ∞ 𝒁 = βˆ’1

πœ”πΆπ‘– β†’ 0

short circuit

𝒁 = πœ”πΏ 𝑖 β†’ ∞open circuit

Circuit response depends on the frequency

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