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Basic Electrical Technology
[ELE 1051]
SINGLE PHASE AC CIRCUITS
L15 – AC Representation & Response
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 1
Topics covered…
 Representation of AC
o Mathematical form
o Graphical form
o Phasors
 AC response of
o Pure Resistor
o Pure Inductor
o Pure Capacitor
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 2
Representing AC
 Consider three sinusoidal signals 𝑥 𝑡 , 𝑦 𝑡 & 𝑧(𝑡) with same frequency
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 3
𝒙(𝒕) = 𝑿𝒎𝒔𝒊𝒏(𝝎𝒕)
𝒚(𝒕) = 𝒀𝒎 𝐬𝐢𝐧 𝝎𝒕 − ∅
𝒛(𝒕) = 𝒁𝒎 𝐬𝐢𝐧 𝝎𝒕 + 𝜽
Graphical Representation
Mathematical Representation
Phasor Representation
Amplitude
→
Time →
𝒚(𝒕)
𝒙(𝒕)
𝜽
∅
𝒛(𝒕)
𝑿𝒓𝒎𝒔
𝒀𝒓𝒎𝒔
𝒁𝒓𝒎𝒔
𝜽
∅
 Representing the relationship between sinusoidal
signals with same frequency in graphical or
mathematical form is tedious
 Phasor representation is often used
R circuit response with AC supply
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 4
𝑣(𝑡) = 𝑉
𝑚 sin 𝜔𝑡
𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡)
𝑽𝒓𝒎𝒔
𝑰𝒓𝒎𝒔
Amplitude
→
Time →
𝒗(𝒕)
𝐢(𝒕)
f = 50 Hz
𝒗(𝒕)
𝒊(𝒕)
𝑹
Graphical Representation Mathematical Representat
Phasor Representation
𝑖 𝑡 =
𝑣(𝑡)
𝑅
‘Current through the resistor
is in phase with the voltage across it’
𝑃 𝑡
Power Associated - Pure Resistive Circuit
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 5
0.02 0.04
𝑣 𝑡
0.06
𝑖 𝑡
Amplitude
Time(sec)
Instantaneous power,
𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) = 𝑉
𝑚𝐼𝑚𝑠𝑖𝑛2𝜔𝑡 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 =
𝟏
𝑻
𝟎
𝑻
𝒑 𝒕 𝒅𝒕
𝑷𝒂𝒗𝒈 =
𝑽𝒎𝑰𝒎
𝟐
= 𝑽𝒓𝒎𝒔𝑰𝒓𝒎𝒔 =
𝑽𝒓𝒎𝒔
𝟐
𝑹
= 𝑰𝒓𝒎𝒔
𝟐 𝑹
L circuit response with AC supply
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 6
𝑣(𝑡) = 𝑉
𝑚 sin 𝜔𝑡
𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡 − 90°)
𝑽𝒓𝒎𝒔
𝑰𝒓𝒎𝒔
Graphical Representation Mathematical Representat
Phasor Representation
f = 50 Hz
𝒗(𝒕)
𝒊(𝒕)
𝑳
𝒗(𝒕)
𝐢(𝒕)
Amplitude
→
Time →
𝑖 𝑡 =
1
𝐿
𝑣 𝑡 𝑑𝑡
‘Current through the inductor lags the
voltage across it by 90°’
𝑉 = 𝑉∠0° 𝐼 = 𝐼∠ − 90°
𝑽
𝑰
=
𝑽∠𝟎°
𝑰∠ − 𝟗𝟎°
= 𝒋𝑿𝑳 𝑤ℎ𝑒𝑟𝑒
𝑉
𝐼
= 𝑋𝐿
𝑿𝑳 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑰𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒆 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆
𝑃 𝑡
Power Associated – Pure Inductive Circuit
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 7
𝑖 𝑡
Amplitude
Time(sec)
0.04 0.06
𝑣 𝑡
0.02
Instantaneous power,
𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 =
𝟏
𝑻
𝟎
𝑻
𝒑 𝒕 𝒅𝒕
𝑷𝒂𝒗𝒈 = 𝟎
= 𝑉
𝑚𝐼𝑚 sin 𝜔𝑡 . sin(𝜔𝑡 − 90 °)
= −
𝑉
𝑚𝐼𝑚
2
sin 2𝜔𝑡
C circuit response with AC supply
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 8
𝑣(𝑡) = 𝑉
𝑚 sin 𝜔𝑡
𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡 + 90°)
𝑽𝒓𝒎𝒔
𝑰𝒓𝒎𝒔
Graphical Representation Mathematical Representat
f = 50 Hz
𝒗(𝒕)
𝒊(𝒕)
𝑪
Phasor Representation
Amplitude
→
Time →
𝒗(𝒕)
𝐢(𝒕)
‘Current through the capacitor
leads the voltage across it by 90°’
𝑖 𝑡 = 𝐶
𝑑𝑣(𝑡)
𝑑𝑡
𝑉 = 𝑉∠0° 𝐼 = 𝐼∠90°
𝑽
𝑰
=
𝑽∠𝟎°
𝑰∠𝟗𝟎°
= −𝒋𝑿𝑪 𝑤ℎ𝑒𝑟𝑒
𝑉
𝐼
= 𝑋𝐶
𝑿𝑪 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆
𝑃 𝑡
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 9
𝑣 𝑡
0.02
Time(sec)
Amplitude
0.04 0.06
Power Associated – Pure capacitive Circuit
𝑖 𝑡
Instantaneous power,
𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 =
𝟏
𝑻
𝟎
𝑻
𝒑 𝒕 𝒅𝒕
𝑷𝒂𝒗𝒈 = 𝟎
= 𝑉
𝑚𝐼𝑚 sin 𝜔𝑡 . sin(𝜔𝑡 + 90 °)
=
𝑉
𝑚𝐼𝑚
2
sin 2𝜔𝑡
Summary
 Sinusoidal alternating signals of same frequency can be
represented graphically by Phasors
 Define: Inductive and capacitive Reactances
Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 10
R L C
Voltage, current
relationship
v(t) in phase with I(t) i(t) lags v(t) by
90°
i(t) leads v(t) by
90°
Power associated
𝐼2𝑅 =
𝑉2
𝑅
(Active Power)
𝐼2
𝑋𝐿
(Reactive Power)
𝐼2
𝑋𝐶
(Reactive Power)

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L15 - AC Representation _ Response.pptx

  • 1. Basic Electrical Technology [ELE 1051] SINGLE PHASE AC CIRCUITS L15 – AC Representation & Response Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 1
  • 2. Topics covered…  Representation of AC o Mathematical form o Graphical form o Phasors  AC response of o Pure Resistor o Pure Inductor o Pure Capacitor Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 2
  • 3. Representing AC  Consider three sinusoidal signals 𝑥 𝑡 , 𝑦 𝑡 & 𝑧(𝑡) with same frequency Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 3 𝒙(𝒕) = 𝑿𝒎𝒔𝒊𝒏(𝝎𝒕) 𝒚(𝒕) = 𝒀𝒎 𝐬𝐢𝐧 𝝎𝒕 − ∅ 𝒛(𝒕) = 𝒁𝒎 𝐬𝐢𝐧 𝝎𝒕 + 𝜽 Graphical Representation Mathematical Representation Phasor Representation Amplitude → Time → 𝒚(𝒕) 𝒙(𝒕) 𝜽 ∅ 𝒛(𝒕) 𝑿𝒓𝒎𝒔 𝒀𝒓𝒎𝒔 𝒁𝒓𝒎𝒔 𝜽 ∅  Representing the relationship between sinusoidal signals with same frequency in graphical or mathematical form is tedious  Phasor representation is often used
  • 4. R circuit response with AC supply Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 4 𝑣(𝑡) = 𝑉 𝑚 sin 𝜔𝑡 𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡) 𝑽𝒓𝒎𝒔 𝑰𝒓𝒎𝒔 Amplitude → Time → 𝒗(𝒕) 𝐢(𝒕) f = 50 Hz 𝒗(𝒕) 𝒊(𝒕) 𝑹 Graphical Representation Mathematical Representat Phasor Representation 𝑖 𝑡 = 𝑣(𝑡) 𝑅 ‘Current through the resistor is in phase with the voltage across it’
  • 5. 𝑃 𝑡 Power Associated - Pure Resistive Circuit Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 5 0.02 0.04 𝑣 𝑡 0.06 𝑖 𝑡 Amplitude Time(sec) Instantaneous power, 𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) = 𝑉 𝑚𝐼𝑚𝑠𝑖𝑛2𝜔𝑡 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 = 𝟏 𝑻 𝟎 𝑻 𝒑 𝒕 𝒅𝒕 𝑷𝒂𝒗𝒈 = 𝑽𝒎𝑰𝒎 𝟐 = 𝑽𝒓𝒎𝒔𝑰𝒓𝒎𝒔 = 𝑽𝒓𝒎𝒔 𝟐 𝑹 = 𝑰𝒓𝒎𝒔 𝟐 𝑹
  • 6. L circuit response with AC supply Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 6 𝑣(𝑡) = 𝑉 𝑚 sin 𝜔𝑡 𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡 − 90°) 𝑽𝒓𝒎𝒔 𝑰𝒓𝒎𝒔 Graphical Representation Mathematical Representat Phasor Representation f = 50 Hz 𝒗(𝒕) 𝒊(𝒕) 𝑳 𝒗(𝒕) 𝐢(𝒕) Amplitude → Time → 𝑖 𝑡 = 1 𝐿 𝑣 𝑡 𝑑𝑡 ‘Current through the inductor lags the voltage across it by 90°’ 𝑉 = 𝑉∠0° 𝐼 = 𝐼∠ − 90° 𝑽 𝑰 = 𝑽∠𝟎° 𝑰∠ − 𝟗𝟎° = 𝒋𝑿𝑳 𝑤ℎ𝑒𝑟𝑒 𝑉 𝐼 = 𝑋𝐿 𝑿𝑳 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑰𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒆 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆
  • 7. 𝑃 𝑡 Power Associated – Pure Inductive Circuit Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 7 𝑖 𝑡 Amplitude Time(sec) 0.04 0.06 𝑣 𝑡 0.02 Instantaneous power, 𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 = 𝟏 𝑻 𝟎 𝑻 𝒑 𝒕 𝒅𝒕 𝑷𝒂𝒗𝒈 = 𝟎 = 𝑉 𝑚𝐼𝑚 sin 𝜔𝑡 . sin(𝜔𝑡 − 90 °) = − 𝑉 𝑚𝐼𝑚 2 sin 2𝜔𝑡
  • 8. C circuit response with AC supply Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 8 𝑣(𝑡) = 𝑉 𝑚 sin 𝜔𝑡 𝑖(𝑡) = 𝐼𝑚sin(𝜔𝑡 + 90°) 𝑽𝒓𝒎𝒔 𝑰𝒓𝒎𝒔 Graphical Representation Mathematical Representat f = 50 Hz 𝒗(𝒕) 𝒊(𝒕) 𝑪 Phasor Representation Amplitude → Time → 𝒗(𝒕) 𝐢(𝒕) ‘Current through the capacitor leads the voltage across it by 90°’ 𝑖 𝑡 = 𝐶 𝑑𝑣(𝑡) 𝑑𝑡 𝑉 = 𝑉∠0° 𝐼 = 𝐼∠90° 𝑽 𝑰 = 𝑽∠𝟎° 𝑰∠𝟗𝟎° = −𝒋𝑿𝑪 𝑤ℎ𝑒𝑟𝑒 𝑉 𝐼 = 𝑋𝐶 𝑿𝑪 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆
  • 9. 𝑃 𝑡 Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 9 𝑣 𝑡 0.02 Time(sec) Amplitude 0.04 0.06 Power Associated – Pure capacitive Circuit 𝑖 𝑡 Instantaneous power, 𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡) 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑷𝒐𝒘𝒆𝒓, 𝑷 = 𝟏 𝑻 𝟎 𝑻 𝒑 𝒕 𝒅𝒕 𝑷𝒂𝒗𝒈 = 𝟎 = 𝑉 𝑚𝐼𝑚 sin 𝜔𝑡 . sin(𝜔𝑡 + 90 °) = 𝑉 𝑚𝐼𝑚 2 sin 2𝜔𝑡
  • 10. Summary  Sinusoidal alternating signals of same frequency can be represented graphically by Phasors  Define: Inductive and capacitive Reactances Monday, March 11, 2024 Dept. of Electrical & Electronics Engg., MIT - Manipal 10 R L C Voltage, current relationship v(t) in phase with I(t) i(t) lags v(t) by 90° i(t) leads v(t) by 90° Power associated 𝐼2𝑅 = 𝑉2 𝑅 (Active Power) 𝐼2 𝑋𝐿 (Reactive Power) 𝐼2 𝑋𝐶 (Reactive Power)

Editor's Notes

  1. Derive in class v(t) and i(t) for L to find XL.
  2. Derive in class v(t) and i(t) for C to find XC.