CHAPTER -7
ALTERNATE CURRENT
➢ PURE AC RESISTOR CIRCUIT OR AC CIRCUIT CONTAINNING ONLY A
RESISTOR:
➢
A Resistor of resistance R is connected in series to a source of alternating emf
𝜀 = 𝜀𝑚 sin 𝜔𝑡 →→→→→→→→→→→→eq1
i= instantaneous current at an instant of time t
acc. To Kirchoff’s law: instantaneous potential drop across resistor = Instantaneous emf of source
VR= ℇ
𝑖𝑅 = 𝜀𝑚 sin 𝜔𝑡
𝜀𝑚
𝑖= sin 𝜔𝑡
𝑅
𝑖 = 𝑖𝑚 sin 𝜔𝑡→→→→→→→→→→→→eq2
𝜺𝒎
𝑤ℎ𝑒𝑟𝑒 𝒊𝒎 =
𝑹
From eq1 and eq2, alternating current and voltage in pure ac resistor circuit are in phase and phase
difference between ε and I is zero
➢ AVERAGE POWER IN PURE AC RESISTOR CIRCUIT:
In pure ac resistor circuit, 𝜀 = 𝜀𝑚 sin 𝜔𝑡 & 𝑖 = 𝑖𝑚 sin 𝜔𝑡
Instantaneous power of pure ac resistor circuit is
𝑃 = 𝑉𝐼 = 𝜀𝑖 = (𝜀𝑚 sin 𝜔𝑡)(𝑖𝑚 sin 𝜔𝑡) = 𝑖𝑚 𝜀𝑚 𝑆𝑖𝑛2 𝜔𝑡
〈𝑷〉 = average power of pure ac resistor circuit
〈𝑃〉 = 〈𝑖𝑚 𝜀𝑚 𝑆𝑖𝑛2 𝜔𝑡〉 = 𝑖𝑚 𝜀𝑚 〈𝑆𝑖𝑛2 𝜔𝑡 〉
We know that 〈𝑆𝑖𝑛2 𝜔𝑡 〉 over a half cycle or full cycle is ½. Therefore,
𝒊𝒎 𝜺𝒎 𝒊𝒎 𝜺𝒎 𝒊𝟐𝒎 𝑹 𝜺𝟐𝒎
〈𝑷〉 = = ( ) ( ) = 𝒊𝒓𝒎𝒔 𝜺𝒓𝒎𝒔 = =
𝟐 √𝟐 √𝟐 𝟐 𝟐𝑹
➢ PURE AC INDUCTOR CIRCUIT OR AC CIRCUIT CONTAINING ONLY AN
INDCUTOR:
An inductor of inductance L is connected in series to an ac source of emf
𝜀 = 𝜀𝑚 sin 𝜔𝑡 →→→ eq1
𝑑𝑖
As alternating current flows through inductor, a back emf −𝐿 𝑑𝑡 is produced which opposes
applied emf.
using Kirchhoff’s loop rule ∑ 𝜀 = ∑ 𝐼𝑅
𝑑𝑖
𝜀−𝐿 =0 ∵𝑅=0
𝑑𝑡
𝑑𝑖
𝜀=𝐿
𝑑𝑡
𝑑𝑖
𝜀𝑚 sin 𝜔𝑡 = 𝐿
𝑑𝑡
𝜀𝑚
𝑑𝑖 = sin 𝜔𝑡 𝑑𝑡
𝐿
𝜀𝑚
∫ 𝑑𝑖 = ∫ sin 𝜔𝑡 𝑑𝑡
𝐿
𝜀𝑚 cos 𝜔𝑡
𝑖= (– )
𝐿 𝜔
𝜀𝑚 𝜀𝑚 𝜋 𝜀𝑚 𝜋
𝑖=− cos 𝜔𝑡 = − sin ( − 𝜔𝑡) = sin (𝜔𝑡 − )
𝐿𝜔 𝐿𝜔 2 𝐿𝜔 2
𝜀𝑚 𝜋
𝑖= sin (𝜔𝑡 − )
𝐿𝜔 2
𝝅
𝒊 = 𝒊𝒎 𝐬𝐢𝐧 (𝝎𝒕 − 𝟐 ) =instantaneous current in pure ac inductor circuit →→→ eq2
So, from eq1 and eq2, in pure ac inductor circuit, the Phase difference between ac voltage and
𝝅
current in ac inductor circuit is 𝟗𝟎𝟎 𝒐𝒓 𝒓𝒂𝒅
𝟐
This indicates that the current lags behind voltage or voltage leads current by a phase angle of
𝜋
900 𝑜𝑟 .
2
𝜺𝒎 𝜺𝒎 𝜺𝒎
𝑾𝒉𝒆𝒓𝒆 𝒊𝒎 = = =
𝑳𝝎 𝑳𝟐𝝅𝒇 𝑿𝑳
Comparing this equation with Ohms law
𝑿𝑳 = 𝑳𝝎 = 𝟐𝝅𝒇𝑳 = inductive reactance is the effective resistance offered by inductor for flow
of ac through it.
𝑤ℎ𝑒𝑟𝑒 𝜔 = 2𝜋𝑓 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑐
𝑓 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
SI unit of inductive reactance is ohm(Ω)
➢ VARIATION OF XL WITH AC FREQUENCY:
𝐹𝑜𝑟 𝑎𝑐 𝑋𝐿 ∝ 𝜔 ∝ 𝑓
So, the graph of XL versus ω or f is a straight line with positive slope
𝐹𝑜𝑟 𝑑𝑖𝑟𝑒𝑐𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡(𝑑𝑐) 𝑓 = 0 𝑆𝑜 𝑋𝐿 = 0. Thus, inductor allows dc flow through it and
blocks the flow of ac through it.
𝒊𝒎 𝜺𝒎 𝜺𝒓𝒎𝒔 𝜺𝒎
𝑭𝒐𝒓 𝒂𝒄 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒊𝒓𝒎𝒔 = = = =
√𝟐 𝑳𝝎√𝟐 𝑳𝝎 𝑿𝑳
➢ PHASE RELATIONSHIP BETWEEN AC VOLTAGE AND CURRENT IN AC
INDUCTOR CIRCUIT:
𝜋
𝐼𝑛 𝑎𝑐 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝜀 = 𝜀𝑚 sin 𝜔𝑡 & 𝑖 = 𝑖𝑚 sin (𝜔𝑡 − )
2
➢ INSTANTANIOUS POWER AND AVERGE POWER OF INDUCTOR IN PURE AC
INDUCTOR CIRCUIT:
𝜋
𝐼𝑛 𝑝𝑢𝑟𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ∶ 𝜀 = 𝜀𝑚 sin 𝜔𝑡 & 𝑖 = 𝑖𝑚 sin (𝜔𝑡 − )
2
𝑰𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒊𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓 𝒐𝒇 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓
𝜋
𝑃 = 𝑉𝑖 = 𝜀𝑖 = 𝜀𝑚 sin 𝜔𝑡 𝑖𝑚 sin (𝜔𝑡 − )
2
𝜋 𝜀𝑚 𝑖𝑚
𝑃 = −𝜀𝑚 𝑖𝑚 sin 𝜔𝑡 sin ( − 𝜔𝑡) = −𝜀𝑚 𝑖𝑚 sin 𝜔𝑡 cos 𝜔𝑡 = − 2 sin 𝜔𝑡 cos 𝜔𝑡
2 2
𝜀𝑚 𝑖𝑚
𝑃=− sin 2𝜔𝑡 = −𝜀𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 sin 2𝜔𝑡
2
〈𝑷〉 = 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 𝒑𝒐𝒘𝒆 𝒐𝒇 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓
𝜀𝑚 𝑖𝑚 𝜀𝑚 𝑖𝑚
〈𝑃〉 = 〈− sin 2𝜔𝑡 〉 = − 〈sin 2𝜔𝑡 〉
2 2
𝐵𝑢𝑡 〈sin 2𝜔𝑡 〉 = 𝑜
∴ 〈𝑷〉 = 𝟎 , Thus, average power of inductor in pure ac inductor circuit is zero
➢ POWER AND ENERGY CHANGES IN INDUCTOR CONNECTED TO AC:
When an inductor of inductance (L) is connected to an ac source it gets magnetised and
demagnetised alternately during each half cycle of alternating current.
1. FIRST QUARTER CYCLE OF AC: AC Current an ac voltage and power is positive. Energy
is absorbed from ac source as core of inductor gets magnetised during first quarter cycle of ac.
➢ 2. SECOND QUARTER CYCLE OF AC: AC current is positive; ac voltage is negative hence
power is negative. Energy is returned to the ac source as the core of inductor gets demagnetised
during second quarter cycle of ac.
➢ 3. THIRD QUARTER CYCLE OF AC: AC current and ac voltage both are negative and hence
power is positive. Energy is absorbed from ac source as core of inductor gets magnetised during
third quarter cycle of ac.
➢ 4. FOURTH QUARTER CYCLE OF AC: AC current is negative; ac voltage is positive hence
power is negative. Energy is returned to the ac source as core of inductor gets demagnetised during
fourth quarter cycle of ac.
➢ PURE CAPACITOR CIRCUIT (OR) AC CIRCUIT CONTAINING ONLY A
CAPACITOR:
A capacitor of capacitance C is connected to an ac source of emf
𝑉 = 𝜀 = 𝜀𝑚 sin 𝜔𝑡
𝑞
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑐𝑒 = 𝐶 =
𝑉
𝑞 = 𝐶𝑉 = 𝐶𝜀𝑚 sin 𝜔𝑡
𝑑𝑞 𝑑(𝐶𝜀𝑚 sin 𝜔𝑡)
𝑖= = = 𝐶𝜔𝜀𝑚 cos 𝜔𝑡
𝑑𝑡 𝑑𝑡
𝜋
𝑖 = 𝐶𝜔𝜀𝑚 cos 𝜔𝑡 = 𝐶𝜔𝜀𝑚 sin ( + 𝜔𝑡)
2
𝜋
𝑖 = 𝐶𝜔𝜀𝑚 sin (𝜔𝑡 + )
2
𝝅
𝒊 = 𝒊𝒎 𝐬𝐢𝐧 (𝝎𝒕 + )
𝟐
Thus, in pure ac capacitor circuit current leads voltage or voltage lags behind current in phase by
𝜋
𝑟𝑎𝑑 𝑜𝑟 900
2
𝜀𝑚 𝜀𝑚
𝑤ℎ𝑒𝑟𝑒 𝒊𝒎 = 𝑪𝝎𝜺𝒎 = =
1⁄ 𝑋𝑐
𝐶𝜔
𝜺𝒎
∴ 𝒊𝒎 =
𝑿𝒄
𝑪𝒐𝒎𝒑𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒊𝒔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒘𝒊𝒕𝒉 𝑶𝒉𝒎𝒔 𝒍𝒂𝒘
𝟏 𝟏
𝑿𝒄 = = = 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆
𝑪𝝎 𝟐𝝅𝒇𝑪
𝑺𝑰 𝒖𝒏𝒊𝒕 𝒐𝒇 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆 𝒊𝒔 𝒐𝒉𝒎(𝛀)
➢ PHASE RELATIONSHIP BETWEEN AC VOLTAGE AND CURRENT IN AC
CAPACITOR CIRCUIT:
𝜋
In ac CAPACITOR circuit: 𝜀 = 𝜀𝑚 sin 𝜔𝑡 & 𝑖 = 𝑖𝑚 sin (𝜔𝑡 + 2 )
So, in pure ac capacitor circuit the current leads voltage or voltage lags behind current by a phase
𝜋
angle of 900 𝑜𝑟 𝑟𝑎𝑑. Phase difference between ac voltage and current in ac capacitor circuit
2
𝜋
is 900 𝑜𝑟 𝑟𝑎𝑑.
2
➢ CAPACITIVE REACTANCE (XC):
CAPACITIVE REACTANCE (XC) is the effective resistance or opposition offered by the
capacitor to the flow of alternating current through it
𝟏 𝟏
𝑿𝒄 = =
𝑪𝝎 𝟐𝝅𝒇𝑪
𝑤ℎ𝑒𝑟𝑒 𝜔 = 2𝜋𝑓 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑐
𝑓 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝟏 𝟏
𝑭𝒐𝒓 𝒂𝒄 𝑿𝒄 ∝ ∝
𝝎 𝒇
𝑭𝒐𝒓 𝒅𝒊𝒓𝒆𝒄𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕(𝒅𝒄) 𝒇 = 𝟎 𝑺𝒐 𝑿𝒄 = ∞
Thus, capacitor allows ac flow through it and offers infinite resistance to the flow of dc through
it.
𝒊𝒎 𝜺𝒎 𝜺𝒓𝒎𝒔 𝜺𝒓𝒎𝒔
𝑭𝒐𝒓 𝒂𝒄 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒊𝒓𝒎𝒔 = = = =
√𝟐 𝟏 𝟏 𝑿𝒄
√𝟐 (𝑪𝝎) 𝑪𝝎
𝑺𝑰 𝒖𝒏𝒊𝒕 𝒐𝒇 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒄𝒆 𝒊𝒔 𝒐𝒉𝒎(𝛀)
➢ VARIATION OF XC WITH AC FREQUENCY:
𝟏 𝟏
𝑭𝒐𝒓 𝒂𝒄 𝑿𝒄 ∝ ∝
𝝎 𝒇
𝑺𝒐 𝒕𝒉𝒆 𝒈𝒓𝒂𝒑𝒉 𝒐𝒇 𝑿𝒄 𝒗𝒆𝒓𝒔𝒖𝒔 𝝎 𝒐𝒓 𝒇 𝒊𝒔 𝒏𝒐𝒕 𝒂 𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆
➢ INSTANTANIOUS POWER AND AVERGE POWER OF CAPACITOR IN PURE AC
CAPACITOR CIRCUIT:
𝐼𝑛 𝑝𝑢𝑟𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝜀 = 𝜀𝑚 sin 𝜔𝑡
𝜋
𝑖 = 𝑖𝑚 sin (𝜔𝑡 + )
2
𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑖𝑜𝑢𝑠 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝑃 = 𝑉𝑖
𝜋
𝑃 = 𝜀𝑖 = 𝜀𝑚 sin 𝜔𝑡 𝑖𝑚 sin (𝜔𝑡 + )
2
𝜋
𝑃 = 𝜀𝑚 𝑖𝑚 sin 𝜔𝑡 sin ( + 𝜔𝑡) = 𝜀𝑚 𝑖𝑚 sin 𝜔𝑡 cos 𝜔𝑡
2
𝜀𝑚 𝑖𝑚
𝑃= 2 sin 𝜔𝑡 cos 𝜔𝑡
2
𝜺𝒎 𝒊𝒎
𝑷= 𝐬𝐢𝐧 𝟐𝝎𝒕 = 𝑰𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓 𝒐𝒇 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓
𝟐
𝑷 = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔 𝐬𝐢𝐧 𝟐𝝎𝒕 = 𝑰𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓 𝒐𝒇 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓
〈𝑃〉 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒 𝑜𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟
𝜀𝑚 𝑖𝑚 𝜀𝑚 𝑖𝑚
〈𝑃〉 = 〈 sin 2𝜔𝑡 〉 = 〈sin 2𝜔𝑡 〉
2 2
𝐵𝑢𝑡 〈sin 2𝜔𝑡 〉 = 𝑜
∴ 〈𝑷〉 = 𝟎 Thus average power of capacitor in pure ac capacitor circuit is zero
➢ POWER AND ENERGY CHANGES IN CAPACITOR CONNECTED TO AC:
When a capacitor is connected to an ac source it gets charged and discharged alternately during
each half cycle of alternating current
1. FIRST QUARTER CYCLE OF AC: AC Current an ac voltage and power is positive. Energy
is absorbed from ac source as capacitor is charged during first quarter cycle of ac.
2. SECOND QUARTER CYCLE OF AC: AC current is negative; ac voltage is positive hence
power is negative. Energy is returned to the ac source as the capacitor gets discharged during
second quarter cycle of ac.
3. THIRD QUARTER CYCLE OF AC: AC current and ac voltage both are negative and hence
power is positive. Energy is absorbed from ac source as capacitor is charged during third quarter
cycle of ac.
4. FOURTH QUARTER CYCLE OF AC: AC current is positive; ac voltage is negative hence
power is negative. Energy is returned to the ac source as the capacitor gets discharged during
fourth quarter cycle of ac.
➢ AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT:
A resistor of resistance R, an inductor of inductance L and a capacitor of capacitance C are
connected in series to an alternating emf
𝜀 = 𝜀𝑚 sin 𝜔𝑡
𝑖 = 𝑖𝑚 sin(𝜔𝑡 + 𝜙) = electric current in series LCR ac circuit at any instant
𝜙 = phase difference between voltage and current
We want to the instantaneous current (i) and its phase relationship to the applied alternating
voltage.
PHASOR DIAGRAM SOLUTION:
The instantaneous current and its phase relationship to the applied ac voltage is obtained from
phasor diagram solution
⃗ 𝑅, 𝑉
𝐿𝑒𝑡 𝑉 ⃗ 𝐿 𝑎𝑛𝑑 𝑉
⃗ 𝐶 represent voltage phasors across resistor, inductor and capacitor respectively
The length or amplitude of these voltage phasors is 𝑉𝑅𝑚 = 𝑅𝑖𝑚 = is voltage amplitude across
resistor in phase with current im.
𝜋
𝑉𝐿𝑚 = 𝑋𝐿 𝑖𝑚 = is voltage amplitude across inductor ahead of current amplitude im by phase 2 .
𝑉𝐶𝑚 = 𝑋𝑐 𝑖𝑚 = is voltage amplitude across capacitor lagging behind current amplitude im by
𝜋
phase 2
⃗ 𝐿 𝑎𝑛𝑑 𝑉
𝑉 ⃗ 𝐶 are in opposite direction whose resultant is ( 𝑉
⃗𝐶 − 𝑉
⃗ 𝐿 ). The resultant of
⃗ 𝑅 𝑎𝑛𝑑 ( 𝑉
𝑉 ⃗𝐶 − 𝑉
⃗ 𝐿 ) is equal to the applied emf.
𝐹𝑟𝑜𝑚 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝑙𝑎𝑤
2 2
⃗ 𝑅) + ( 𝑉
𝜀2 = ( 𝑉 ⃗𝐶 − 𝑉
⃗ 𝐿)
𝜀 2 = ( 𝑅𝑖𝑚 )2 + ( 𝑋𝐶 𝑖𝑚 − 𝑋𝐿 𝑖𝑚 )2
𝜀 2 = [𝑅 2 + (𝑋𝐶 − 𝑋𝐿 )2 ]𝑖𝑚
2
𝜺 𝜺
𝒊𝒎 = =
√𝑹𝟐 + (𝑿𝑪 − 𝑿𝑳 )𝟐 𝒁
𝒁 = √𝑹𝟐 + (𝑿𝑪 − 𝑿𝑳 )𝟐 is called impedance(Z) which is the effective resistance of series LCR
ac circuit.
The relationship between 𝑉𝑅 , (𝑉𝐶 − 𝑉𝐿 ) and Z or relatioship between 𝑋𝑅 , (𝑋𝐶 − 𝑋𝐿 ) 𝑎𝑛𝑑 𝑍 is
obtained from Impedance diagram or impedance triangle which is a right angled triangle with Z
as hypotenuse 𝑉𝑅 𝑎𝑛𝑑 (𝑉𝑅 − 𝑉𝑅 ) as adjacent sides.
SPECIAL CASES:
𝐶𝐴𝑆𝐸 − 𝐼: 𝑾𝒉𝒆𝒏 𝑽𝑪 < 𝑽𝑳 𝒐𝒓𝑿𝑪 < 𝑿𝑳
𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑙𝑎𝑔𝑠 𝑏𝑒ℎ𝑖𝑛𝑑 𝑜𝑓 𝑒𝑚𝑓 𝑏𝑦
𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑎𝑛𝑑 𝒕𝒉𝒆 𝑳𝑪𝑹 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
𝒊𝒔 𝒔𝒂𝒊𝒅 𝒕𝒐 𝒃𝒆 𝒊𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒆
𝑉𝐿 − 𝑉𝐶 𝑋𝐿 − 𝑋𝐶
tan 𝜙 = =
𝑅 𝑅
𝑖 = 𝑖𝑚 sin(𝜔𝑡 − 𝜙)
𝐶𝐴𝑆𝐸 − 2: 𝑾𝒉𝒆𝒏 𝑽𝑪 > 𝑽𝑳 𝒐𝒓𝑿𝑪 > 𝑿𝑳 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑙𝑒𝑎𝑑𝑠 𝑜𝑟 𝑎ℎ𝑒𝑎𝑑 𝑜𝑓 𝑒𝑚𝑓 𝑏𝑦
𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑎𝑛𝑑 𝒕𝒉𝒆 𝑳𝑪𝑹 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒊𝒔 𝒔𝒂𝒊𝒅 𝒕𝒐 𝒃𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒊𝒗𝒆
𝑉𝐶 − 𝑉𝐿 𝑋𝐶 − 𝑋𝐿
tan 𝜙 = =
𝑅 𝑅
𝑖 = 𝑖𝑚 sin(𝜔𝑡 + 𝜙)
𝐶𝐴𝑆𝐸 − 3: 𝑾𝒉𝒆𝒏 𝑽𝑪 = 𝑽𝑳 𝒐𝒓𝑿𝑪 = 𝑿𝑳 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑛𝑑 𝑒𝑚𝑓 𝑎𝑟𝑒 𝑖𝑛
𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑳𝑪𝑹 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒊𝒔 𝒔𝒂𝒊𝒅 𝒕𝒐 𝒃𝒆 𝒑𝒖𝒓𝒆𝒍𝒚 𝒓𝒆𝒔𝒊𝒔𝒕𝒊𝒗𝒆
𝜀𝑚 𝑖𝑚 𝜀𝑚 𝜀𝑟𝑚𝑠
tan 𝜙 = 0 ; 𝑖𝑚 = ; 𝑖𝑟𝑚𝑠 = = =
𝑍 √2 𝑍√2 𝑍
➢ RESONANCE CONDITION OF A SERIES LCR AC CIRCUIT:
𝐴 𝑠𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑎𝑐 𝑐𝑖𝑟𝑐𝑢𝑡 𝑖𝑠 𝑠𝑎𝑖𝑑 𝑡𝑜 𝑏𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒
𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑖𝑡 ℎ𝑎𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑤ℎ𝑖𝑐ℎ 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛
𝑖𝑚𝑝𝑒𝑑𝑎𝑐𝑒 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 𝑋𝐶 = 𝑋𝐿
𝑚𝑎𝑥
𝑖𝑚 = 𝑖𝑚 𝑤ℎ𝑒𝑛 𝑍 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 𝑋𝐶 = 𝑋𝐿
𝑇ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎𝑐 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑋𝐶 = 𝑋𝐿 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑐𝑦 (𝜔𝑟 )
𝐴𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝜔𝑟 ) 𝑋𝐶 = 𝑋𝐿
1
= 𝜔𝑟 𝐿
𝜔𝑟 𝐶
1
𝜔𝑟2 =
𝐿𝐶
𝟏
𝝎𝒓 = 𝟐𝝅𝒇𝒓 = = 𝒓𝒆𝒔𝒐𝒏𝒂𝒏𝒄𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒐𝒇 𝒔𝒆𝒓𝒊𝒆𝒔 𝑳𝑪𝑹 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
√𝑳𝑪
𝟏
𝒇𝒓 = = 𝒓𝒆𝒔𝒐𝒏𝒂𝒏𝒄𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒐𝒇 𝒔𝒆𝒓𝒊𝒆𝒔 𝑳𝑪𝑹 𝒂𝒄 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
𝟐𝝅√𝑳𝑪
➢ 𝑪𝑯𝑨𝑹𝑨𝑪𝑻𝑬𝑹𝑰𝑺𝑻𝑰𝑪𝑺 𝑶𝑭 𝒔𝒆𝒓𝒊𝒆𝒔 𝑳𝑪𝑹 𝒂𝒄 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
1. 𝑹𝒆𝒔𝒐𝒏𝒂𝒏𝒄𝒆 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝑓𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑎𝑐 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑿𝒄 = 𝑿𝑳
2. 𝒁 = 𝑹 = 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒊𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆
𝜺𝒎
3. 𝒊𝒎𝒂𝒙
𝒎 = = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑚𝑝𝑖𝑡𝑢𝑑𝑒
𝑹
𝟏
4. 𝝎𝒓 = 𝟐𝝅𝒇𝒓 = = 𝒓𝒆𝒔𝒐𝒏𝒂𝒏𝒄𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
√𝑳𝑪
𝜺𝟐𝒎
5. 𝑷 = = 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒐𝒘𝒆𝒓 𝑖𝑠 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝑏𝑦 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑹
𝑖𝑛 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑠𝑡𝑎𝑡𝑒
6. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑅 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑒𝑚𝑓
7. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟(𝐿)𝑎𝑛𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 (𝐶) 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙
𝑎𝑛𝑑 ℎ𝑎𝑣𝑒 𝒑𝒉𝒂𝒔𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝟏𝟖𝟎𝟎 ℎ𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑖𝑟 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑖𝑠 𝑧𝑒𝑟𝑜
8. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟(𝐿)𝑎𝑛𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟(𝐶) 𝑎𝑟𝑒 𝑣𝑒𝑟𝑦 ℎ𝑖𝑔ℎ ℎ𝑒𝑛𝑐𝑒
𝑠𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝑎 𝑙𝑎𝑟𝑔𝑒 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑐 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
9. 𝑆𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑐𝑐𝑒𝑝𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑏𝑒𝑐𝑎𝑢𝑠𝑒
𝑤ℎ𝑒𝑛 𝑙𝑎𝑟𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑎𝑟𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑖𝑡 𝑎𝑐𝑐𝑒𝑝𝑡𝑠 𝑜𝑛𝑙𝑦
𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝑓𝑟 )𝑎𝑛𝑑 𝑟𝑒𝑗𝑒𝑐𝑡𝑠 𝑜𝑡ℎ𝑒𝑟
10. 𝑅𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑜𝑐𝑐𝑢𝑟𝑠 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑐𝑒 𝑅
➢ COMPARISION BETWEEN AC RESISITOR, AC INDUCTOR, AC CAPACITOR AND
SERIES LCR AC CIRCUITS
➢ POWER IN AN AC CIRCUIT:
𝑇ℎ𝑒 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑖𝑛 𝑎𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑝𝑜𝑤𝑒𝑟. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑎𝑛 𝑎𝑐 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑤ℎ𝑜𝑠𝑒 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑖𝑜𝑢𝑠
𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝜀 = 𝜀𝑚 sin 𝜔𝑡
𝑖 = 𝑖𝑚 sin(𝜔𝑡 − 𝜙)
𝑤ℎ𝑒𝑟𝑒 𝜙 𝑖𝑠 𝑡ℎ𝑒 𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑦 𝑤ℎ𝑖𝑐ℎ 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑙𝑒𝑎𝑑𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑎𝑐 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑃 = 𝜀𝑖
𝑃 = 𝜀𝑚 sin 𝜔𝑡 𝑖𝑚 sin(𝜔𝑡 − 𝜙)
𝑃 = 𝜀𝑚 𝑖𝑚 sin 𝜔𝑡 [sin 𝜔𝑡 cos 𝜙 − cos 𝜔𝑡 sin 𝜙]
𝑃 = 𝜀𝑚 𝑖𝑚 [𝑠𝑖𝑛2 𝜔𝑡 cos 𝜙 − sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙]
1 − cos 2𝜔𝑡
𝑃 = 𝜀𝑚 𝑖𝑚 [( ) cos 𝜙 − sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙]
2
𝜀𝑚 𝑖𝑚
𝑃= [(1 − cos 2𝜔𝑡) cos 𝜙 − 2 sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙]
2
𝜀𝑚 𝑖𝑚
𝑃= [cos 𝜙 − (cos 2𝜔𝑡 cos 𝜙 + sin 2𝜔𝑡 sin 𝜙)]
2
𝜺𝒎 𝒊𝒎
𝑷= [𝐜𝐨𝐬 𝝓 − 𝐜𝐨𝐬(𝟐𝝎𝒕 − 𝝓)] = 𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓
𝟐
𝒊𝒏 𝒂𝒄 𝒄𝒊𝒓𝒄𝒖𝒊𝒕
𝑑𝑤 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑖𝑜𝑢𝑠 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑑𝑡
𝑑𝑤 = 𝑃𝑑𝑡
𝑇 𝑇
𝑊 = ∫ 𝑃𝑑𝑡 = ∫ 𝜀𝑖𝑑𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑐𝑦𝑐𝑙𝑒
0 0
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑎 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑐𝑦𝑐𝑙𝑒 𝑜𝑓 𝑎𝑐 〈𝑃〉
𝑇
𝑊 𝜀𝑚 𝑖𝑚
〈𝑃〉 = = ∫[cos 𝜙 − cos(2𝜔𝑡 − 𝜙)]𝑑𝑡
𝑇 2𝑇
0
𝑇 𝑇
𝜀𝑚 𝑖𝑚
〈𝑃〉 = ∫ cos 𝜙 𝑑𝑡 − ∫[cos(2𝜔𝑡 − 𝜙)]𝑑𝑡
2𝑇
0 0
𝜀𝑚 𝑖𝑚 𝜀𝑚 𝑖𝑚
〈𝑃〉 = cos 𝜙 [𝑇]𝑇0 − [cos 𝜙 (𝑇 − 0)]
2𝑇 2𝑇
𝜀𝑚 𝑖𝑚
〈𝑃〉 = cos 𝜙 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑎 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑐𝑦𝑐𝑙𝑒
2
𝜀𝑚 𝑖𝑚
〈𝑃〉 = cos 𝜙
√2 √2
𝜺𝒎 𝒊𝒎 𝑹
〈𝑷〉 = 𝐜𝐨𝐬 𝝓 = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔 𝐜𝐨𝐬 𝝓 = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔
𝟐 𝒁
〈𝑷〉 = 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒑𝒐𝒘𝒆𝒓 𝒐𝒗𝒆𝒓 𝒂 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒆 𝒄𝒚𝒄𝒍𝒆
𝑺𝑷𝑬𝑪𝑰𝑨𝑳 𝑪𝑨𝑺𝑬𝑺 ∶
𝟏. 𝑷𝑼𝑹𝑬 𝑨𝑪 𝑹𝑬𝑺𝑰𝑺𝑻𝑶𝑹 𝑪𝑰𝑹𝑪𝑼𝑰𝑻: 𝐼𝑛 𝑝𝑢𝑟𝑒 𝑎𝑐 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑟𝑒 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝑠𝑜 𝜙 = 0
𝜺𝒎 𝒊𝒎
〈𝑷〉 = = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔
𝟐
𝟐. 𝑷𝑼𝑹𝑬 𝑨𝑪 𝑰𝑵𝑫𝑪𝑼𝑻𝑶𝑹 𝑪𝑰𝑹𝑪𝑼𝑰𝑻: 𝐼𝑛 𝑝𝑢𝑟𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝜋
𝑙𝑒𝑎𝑑𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝑏𝑦 𝜙 =
2
𝜀𝑚 𝑖𝑚 𝜋
〈𝑃〉 = cos ( ) = 0
2 2
𝟑. 𝑷𝑼𝑹𝑬 𝑨𝑪 𝑪𝑨𝑷𝑨𝑪𝑰𝑻𝑶𝑹 𝑪𝑰𝑹𝑪𝑼𝑰𝑻: 𝐼𝑛 𝑝𝑢𝑟𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝜋
𝑙𝑎𝑔𝑠 𝑏𝑒ℎ𝑖𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝑏𝑦 𝜙 =
2
𝜀𝑚 𝑖𝑚 𝜋
〈𝑃〉 = cos ( ) = 0
2 2
𝟒. 𝑺𝑬𝑹𝑰𝑬𝑺 𝑳𝑪𝑹 𝑪𝑰𝑹𝑪𝑼𝑰𝑻: 𝐼𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
ℎ𝑎𝑠 𝑎 𝑝ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜙
𝜀𝑚 𝑖𝑚 𝑋𝐿 − 𝑋𝐶
〈𝑃〉 = cos 𝜙 𝑤ℎ𝑒𝑟𝑒 𝜙 = 𝑡𝑎𝑛− 1 ( )
2 𝑅
𝑆𝑜 𝑝𝑜𝑤𝑒𝑟 𝑖𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑖𝑛 𝑠𝑢𝑐ℎ 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑠 𝑜𝑛𝑙𝑦 𝑖𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟𝑅
𝟓. 𝑹𝑬𝑺𝑶𝑵𝑨𝑵𝑪𝑬 𝑳𝑪𝑹 𝑪𝑰𝑹𝑪𝑼𝑰𝑻: 𝐼𝑛 𝑅𝐸𝑆𝑂𝑁𝐴𝑁𝐶𝐸 𝑆𝐸𝑅𝐼𝐸𝑆 𝐿𝐶𝑅 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑋𝐶 = 𝑋𝐿 𝐴𝑁𝐷 𝜙 = 0 𝐻𝐸𝑁𝐶𝐸 cos 𝜙 = 1
𝜺𝒎 𝒊𝒎
〈𝑷〉 = = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔 = 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒑𝒐𝒘𝒆𝒓 𝒊𝒔 𝒅𝒊𝒔𝒔𝒊𝒑𝒂𝒕𝒆𝒅
𝟐
𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹
➢ POWER FACTOR:
Power factor (Cosφ) is defined as the ratio of true power to the apparent power of an ac
circuit.
〈𝑷〉 = 𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔 𝐜𝐨𝐬 𝝓
𝒕𝒓𝒖𝒆 𝒑𝒐𝒘𝒆𝒓 (〈𝑷〉) = 𝒂𝒑𝒂𝒓𝒆𝒏𝒕 𝒑𝒐𝒘𝒆𝒓(𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔 ) 𝒙 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓 (𝐜𝐨𝐬 𝝓)
〈𝑷〉
𝐜𝐨𝐬 𝝓 = = 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓
𝜺𝒓𝒎𝒔 𝒊𝒓𝒎𝒔
𝑹 𝑹
𝐜𝐨𝐬 𝝓 = = = 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓
𝒁 √𝑹𝟐 + (𝑿𝑪 − 𝑿𝑳 )𝟐
∴ 𝐜𝐨𝐬 𝝓 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓 𝒘𝒉𝒐𝒔𝒆 𝒗𝒂𝒍𝒖𝒆 𝒓𝒂𝒏𝒈𝒆𝒔 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝟎 𝒂𝒏𝒅 𝟏
𝑭𝒐𝒓 𝒑𝒖𝒓𝒆 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓 𝒂𝒏𝒅 𝒑𝒖𝒓𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓 𝒊𝒔 𝟎
𝑭𝒐𝒓 𝒑𝒖𝒓𝒆 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝒑𝒐𝒘𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓 𝒊𝒔 𝟏
➢ WATTLESS CURRENT: The current in ac circuit is said to be wattles if the average power
consumed in the circuit is zero.
The current in purely inductive circuit or purely capacitive circuit in which the voltage and
current differ by a phase angle φ = 900 is called wattless current
〈𝑃〉 = 𝜀𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑎𝑐 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑇ℎ𝑒 𝑟𝑚𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑟𝑚𝑠 𝑖𝑠 𝑖𝑛 𝑡𝑜 𝑡𝑤𝑜 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠𝑖𝑟𝑚𝑠 cos 𝜙 𝑎𝑙𝑜𝑛𝑔 𝜀𝑟𝑚𝑠 𝑎𝑛𝑑
𝑖𝑟𝑚𝑠 sin 𝜙 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝜀𝑟𝑚𝑠
1. 𝐹𝑜𝑟 𝑖𝑟𝑚𝑠 cos 𝜙 𝑎𝑙𝑜𝑛𝑔 𝜀𝑟𝑚𝑠
〈𝑃〉 = 𝜀𝑟𝑚𝑠 (𝑖𝑟𝑚𝑠 cos 𝜙) cos 0 = 𝜀𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
2. 𝐹𝑜𝑟 𝑖𝑟𝑚𝑠 sin 𝜙 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝜀𝑟𝑚𝑠
𝜋
〈𝑃〉 = 𝜀𝑟𝑚𝑠 (𝑖𝑟𝑚𝑠 sin 𝜙) cos = 0
2
𝑯𝒆𝒏𝒄𝒆 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒊𝒓𝒎𝒔 𝐬𝐢𝐧 𝝓 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒘𝒂𝒕𝒕𝒍𝒆𝒔𝒔 𝒄𝒖𝒓𝒓𝒆𝒏𝒕
➢ MAGNETIC ENERGY STORED IN AN INDCUTOR (UB) :
𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑐𝑒 𝑜𝑓 𝑎𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
𝑖 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑎𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
𝑑𝑖
𝜀 = −𝐿 = 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑎𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
𝑑𝑡
𝑑𝑤 = 𝑃𝑑𝑡 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑎𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑑𝑡
𝑑𝑖
𝑑𝑤 = 𝑃𝑑𝑡 = −𝜀𝑖 𝑑𝑡 = − (−𝐿 ) 𝑖 𝑑𝑡 = 𝐿𝑖 𝑑𝑖
𝑑𝑡
𝑖𝑚
𝑊 = ∫ 𝑑𝑤 = 𝑡𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑓𝑜𝑟𝑚 0 𝑡𝑜 𝑖𝑚
0
𝑖𝑚 𝑖𝑚 𝑖𝑚
𝑖
𝑖2 𝑚 1 2
𝑊 = ∫ 𝑑𝑤 = ∫ 𝐿𝑖 𝑑𝑖 = 𝐿 ∫ 𝑖 𝑑𝑖 = 𝐿 [ ] = 𝐿𝑖𝑚
2 0 2
0 0 0
𝑇ℎ𝑒 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑠 𝑠𝑡𝑜𝑟𝑒𝑑 𝑎𝑠 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦(𝑈𝐵 ) 𝑖𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
ℎ𝑒𝑛𝑐𝑒 𝑊 = 𝑈𝐵
𝟏 𝟐
𝑼𝑩 = 𝑳𝒊 = 𝒎𝒂𝒈𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒕𝒐𝒓𝒆𝒅 𝒊𝒏 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓
𝟐 𝒎
➢ TRANSFORMER:
A transformer is an electrical device for converting alternating current at low voltage in to that of
a high voltage or vice versa.
PRINCIPLE OF TRANSFORMER: Transformer works on the principle of mutual induction
CONSTRUCTION OF TRANSFORMER: A transformer consists of two coils of copper wire
wound over soft iron core which provide inductive coupling between coils through magnetic flux.
The coil to which input alternating voltage is
applied is called PRIMARY COIL. The coil
from which output is obtained is called
SECONDARY COIL.
WORKING OF TRANSFORMER: As the
alternating input current flows through the
primary coil it generates alternating
magnetic flux in the core which also passes
through the secondary coil. The alternately
changing magnetic flux produce induced emf in the secondary coil and also a self-induced emf in
the primary coil.
𝐿𝑒𝑡 𝑁𝑝 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑜𝑓 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑖𝑙
𝑁𝑠 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑜𝑓 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑖𝑙
𝜙𝐵 = 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑙𝑢𝑥 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑎𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑖𝑙𝑠
𝑑𝜙𝐵
𝜀𝑝 = −𝑁𝑝 = 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑖𝑙
𝑑𝑡
𝑑𝜙𝐵
𝜀𝑠 = −𝑁𝑠 = 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑖𝑙
𝑑𝑡
𝜀𝑠 𝑁𝑠
= →→→→→→→→→→→ 𝑒𝑞 − 1
𝜀𝑝 𝑁𝑝
𝑖𝑛 𝑖𝑑𝑒𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 (𝜀𝑝 𝑖𝑝 ) = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 (𝜀𝑠 𝑖𝑠 ) →→→→→→→→→→ 𝑒𝑞 − 2
𝑓𝑟𝑜𝑚 𝑒𝑞 − 1 𝑎𝑛𝑑 𝑒𝑞 − 2
𝜺𝒔 𝑵𝒔 𝒊𝒑
= = = 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒓𝒂𝒕𝒊𝒐
𝜺𝒑 𝑵𝒑 𝒊𝒔
𝑇ℎ𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑖𝑠
𝒐𝒖𝒕𝒑𝒖𝒕 𝒑𝒐𝒘𝒆𝒓 (𝑷𝟎 = 𝜺𝒔 𝒊𝒔 )
𝜼= 𝒙𝟏𝟎𝟎
𝒊𝒏𝒑𝒖𝒕 𝒑𝒐𝒘𝒆𝒓(𝑷𝒊 = 𝜺𝒑 𝒊𝒑 )
➢ TYPES OF TRANSFORMERS:
1. STEPUP TRANSFORMER in which output voltage and output power are GREATER THAN
their corresponding input values.
𝑁𝑠 > 𝑁𝑝
𝜀𝑠 > 𝜀𝑝
𝑖𝑠 < 𝑖𝑝
2. STEP DOWN TRANSFORMER in which output voltage and output power are LESSTHAN
than their corresponding input values.
𝑁𝑠 < 𝑁𝑝
𝜀𝑠 < 𝜀𝑝
𝑖𝑠 > 𝑖𝑝
➢ ENERGY LOSSES IN TRANSFORMERS
ENERGY LOSS IN TRANSFORMER METHODS TO MINIMISE ENERGY LOSS
1. COPPER LOSS: Energy is lost due The power loss (P =i2R) can be minimised
to heating of copper wires used in primary by using thick copper wires.
and secondary coils.
2.EDDY CURRENT LOSS: Eddy current loss can be minimised by using
core made up of laminated iron sheets
The alternating magnetic flux induces eddy
currents in the iron core which leads to energy
loss in the form of heat.
3.HYSTERESIS LOSS: The alternating Hysteresis loss can be minimised by using core
current carries the iron core through a cycle of material having narrow hysteresis loop
magnetisation and demagnetisation which
causes loss of energy during each cycle in the
form of heat.
4.FLUX LEAKAGE: The magnetic flux The magnetic flux leakage can be minimised
produced by the primary may not pass fully by winding the primary and secondary coils
through the secondary coil and flux leak in to over one another.
air
5. HUMMING SOUND LOSS: As the core The loss of energy in the form of humming
of transformer goes though each cycle of sound is minimised by immersing the core of
hysteresis loop due to alternating voltage the transformer in oil which absorb vibrations.
core lengthens and shortens due to a
phenomenon called magnetostriction which
produce humming sound.
➢ USE OF TRANSFORMERS IN LONG DISTANCE TRANSMISSION OF
ELECTRICITY:
𝑃
The power supplied by ac generator is P=VI. Since 𝐼 = 𝑉 for given amount of electric power(P),
the power loss is less if I is less or V is high. Hence step up transformers are used at power stations
and electricity is transmitted over long distances at high voltages which minimises power loss
during transmission
❖ IMPORTANT NOTE: Transformer does not work on dc because dc does not change with time
and hence dc passing through a primary cannot change magnetic flux with secondary. So no
output voltage will be produced at the secondary.
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