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Alternating Current
 Alternating Current

• Angle between 𝐵 and 𝐴Ԧ is changing continuously.

𝜔 𝐵
𝜙 = 𝐵𝐴 cos 𝜔𝑡 + 𝜃

⇒ E = 𝐵𝐴𝜔 sin 𝜔𝑡 + 𝜃 𝑖
𝑖
E = E0 sin 𝜔𝑡 + 𝜃 𝜃

𝐵𝐴𝜔
⇒𝑖= sin 𝜔𝑡 + 𝜃
𝑟

𝑖 = 𝑖0 sin 𝜔𝑡 + 𝜃
 Mean Value of a Current in a Sinusoidal AC

Mean value of current for a full cycle:

2𝜋ൗ 2𝜋ൗ
𝜔
‫׬‬0 𝜔 𝑖 𝑑𝑡 ‫׬‬0 𝑖0 sin 𝜔𝑡 𝑑𝑡 𝑖
< 𝑖 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒 = 2𝜋ൗ
= 2𝜋ൗ
𝜔
‫׬‬0 𝜔 𝑑𝑡 ‫׬‬0 𝑑𝑡

𝑡
𝜔𝑖0
2𝜋ൗ
𝜔 𝜋 2𝜋 3𝜋
< 𝑖 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒 = න sin 𝜔𝑡 𝑑𝑡 𝜔 𝜔 𝜔
2𝜋 0

−𝑖0 2𝜋ൗ
𝜔
< 𝑖 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒 = cos 𝜔𝑡 0
2𝜋
𝑖 = 𝑖0 sin 𝜔𝑡
−𝑖0
< 𝑖 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒 = 1−1
2𝜋 𝑖 = Instantaneous value of current

𝑖0 = Peak value of current

< 𝑖 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒 = 0
 Mean Value of a Current in a Sinusoidal AC

Mean value of current for a half cycle:

𝑖
𝜋Τ 𝜋Τ
𝜔 𝜔
‫׬‬0 𝑖 𝑑𝑡 ‫׬‬0 𝑖0 sin 𝜔𝑡 𝑑𝑡
< 𝑖 >ℎ𝑎𝑙𝑓 𝑐𝑦𝑐𝑙𝑒 = 𝜋Τ = 𝜋Τ 𝑡
‫׬‬0 𝜔 𝑑𝑡 ‫׬‬0 𝜔 𝑑𝑡
𝜋 2𝜋 3𝜋
𝜔 𝜔 𝜔
𝜋Τ
𝜔𝑖0 𝜔
< 𝑖 >ℎ𝑎𝑙𝑓 𝑐𝑦𝑐𝑙𝑒 = න sin 𝜔𝑡 𝑑𝑡
𝜋 0
𝑖 = 𝑖0 sin 𝜔𝑡
−𝑖0 𝜋Τ 𝑖 = Instantaneous value of current
𝜔
< 𝑖 >ℎ𝑎𝑙𝑓 𝑐𝑦𝑐𝑙𝑒 = cos 𝜔𝑡 0 𝑖0 = Peak value of current
𝜋

Mean Value of Sinusoidal AC

−𝑖0
< 𝑖 >ℎ𝑎𝑙𝑓 𝑐𝑦𝑐𝑙𝑒 = −1 − 1 Element For Full Cycle For Half Cycle
𝜋
2𝑖𝑜
< 𝑖 > 0
𝜋
2𝑖0 2E𝑜
< 𝑖 >ℎ𝑎𝑙𝑓𝑐𝑦𝑐𝑙𝑒 = <E> 0
𝜋 𝜋
 RMS Value of Sinusoidal AC

In kinetic theory of gases: In alternating current:

Root mean square speed is defined as Root mean square current or voltage is
the square root of mean of squares of defined as the square root of mean
the speed of different molecules. value of square of instantaneous
current or voltage, respectively.

2 2 2
𝑣1 + 𝑣2 + 𝑣3 + 𝑣4 + …2 𝑖𝑟𝑚𝑠 = < 𝑖2 >
𝑣𝑟𝑚𝑠 =
𝑁

E𝑟𝑚𝑠 = < E2 >

 RMS Value of Sinusoidal AC

Mean value of 𝑖 2 :

< 𝑖 2 >ℎ𝑎𝑙𝑓𝑐𝑦𝑐𝑙𝑒 = < 𝑖 2 >𝑓𝑢𝑙𝑙 𝑐𝑦𝑐𝑙𝑒

𝜋Τ
𝑖2
2 2
‫׬‬0 𝜔 𝑖0 2 sin2 𝜔 𝑡 𝑑𝑡
< 𝑖 > = < 𝑖 >ℎ𝑎𝑙𝑓𝑐𝑦𝑐𝑙𝑒 = 𝜋Τ
‫׬‬0 𝜔 𝑑𝑡 𝑡
𝜋 2𝜋 3𝜋
𝜋Τ 𝜔 𝜔 𝜔
‫׬‬0 𝜔
𝑖0 2 sin2 𝜔 𝑡 𝑑𝑡
< 𝑖2 > = 𝜋Τ
‫׬‬0 𝜔 𝑑𝑡

𝜋Τ
2
𝜔𝑖0 2 𝜔
<𝑖 >= න [1 − cos 2𝜔𝑡 ] 𝑑𝑡
2𝜋 0

2
𝜔𝑖0 2 𝜋 1 𝜋Τ 𝑖 0
2
<𝑖 >= − sin 2𝜔𝑡 0
𝜔 < 𝑖 2 >=
2𝜋 𝜔 2𝜔 2
RMS Value of Sinusoidal AC

𝑖0 E0
𝑖𝑟𝑚𝑠 = < 𝑖2 >= E𝑟𝑚𝑠 = < E 2 >=
2 2

• In Indian system, the voltage supply is

220 𝑉, 50 𝐻𝑧.
• The current becomes zero twice in
each cycle, means the current
becomes zero, 100 times in 1 𝑠.
• This 220 𝑉 is the RMS value of the supply
voltage.
• Peak Value supply voltage:
E 0 = E 𝑟𝑚𝑠 2 = 220 2
E 0 = 311 𝑉
• The reading given on the home
appliances is also the RMS value.
 Why RMS ?

𝑡
‫׬‬0 𝑖0 2 sin2 𝜔𝑡 𝑑𝑡
𝑖𝑟𝑚𝑠 2 = 𝑡
….Equation 1
‫׬‬0 𝑑𝑡

Heat energy produced in AC circuit through resistor 𝑅

𝑅 in time 𝑡: 𝑖 = 𝑖0 sin 𝜔𝑡
𝑡
𝐻 = න 𝑖 2 𝑅𝑑𝑡
0
E = E0 sin 𝜔𝑡
𝑡
⇒ 𝐻 = න 𝑖0 2 sin2 𝜔𝑡 𝑅𝑑𝑡 (From equation 1)
0
𝑡
⇒ 𝐻 = 𝑖𝑟𝑚𝑠 2 𝑅 න 𝑑𝑡
0

𝐻 = 𝑖𝑟𝑚𝑠 2 𝑅 𝑡
 Why RMS ?

• Even if we replace the AC voltage

E = E 0 s i n 𝜔 𝑡 by constant voltage
E0
, the heat energy produced 𝑅
2
through resistor 𝑅 in time 𝑡 will still 𝑖 = 𝑖0 sin 𝜔𝑡
be the same.

E = E0 sin 𝜔𝑡
• As it is difficult to understand the
time varying AC for common
people, RMS values are used to
make calculations easier.

𝑖𝑜 𝑅
• The electricity bill at home is
2
calculated with RMS value.

E0
E 𝑟𝑚𝑠 =
2
 Phasor Diagram

A phasor is a vector which rotates about the origin with some angular velocity 𝜔.

• Alternating current and voltage of same frequency are represented as rotating

vectors (phasors) along with proper phase angle between them.
E = E0 sin(𝜔𝑡)

𝑦 𝑖, E 𝒊 = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝜔
𝑖0 E
𝒊
𝑖0 sin 𝜔𝑡1
E 0 sin 𝜔𝑡1 𝜔 𝑡1 E0
𝑥 𝜋 𝜔𝑡
0 𝜔𝑡1 2𝜋

• The vertical projections of phasor represents the sinusoidally varying quantities.

 AC Voltage Applied to a Resistor

Instantaneous current:
𝑅
E0
𝑖(𝑡) = sin(𝜔𝑡 + 𝜙) = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝑅
Relation between peak current and voltage:
E0 E0 sin(𝜔𝑡 + 𝜙)
𝑖0 =
𝑅
Relation between RMS current and voltage:

E𝑟𝑚𝑠
𝑖𝑟𝑚𝑠 =
𝑅
Relation between instantaneous current and voltage:

E( 𝑡 )
𝑖(𝑡) =
𝑅

• Current and Voltage are in phase for purely

resistive circuit
 Find the RMS value of the given current waveform.
T

Solution: RMS value of current (𝑖),

𝑖 (𝐴)
Time period of 𝑖, 𝑇 = 4 𝑠
4
Variation of current (𝑖) with 1 50
𝑖𝑟𝑚𝑠 = න 12.5𝑡 2 𝑑𝑡
time (𝑡) 4
0
50
𝑖= 𝑡 = 12.5𝑡
4 0 4 8 12 𝑡 (𝑠)
4
(12.5)2 𝑡 3
RMS value of current (𝑖), =
4 3 0
𝑇
1
𝑖𝑟𝑚𝑠 = න 𝑖 2 𝑑𝑡 12.5 × 4
𝑇 =
0
3
4
1
= න 12.5𝑡 2 𝑑𝑡 50
4 𝑖𝑟𝑚𝑠 = 𝐴
0 3
 AC Voltage Applied to a Capacitor

Capacitive reactance:
1 1
𝑋𝐶 = = Ω
𝜔𝐶 2𝜋𝑓𝐶

Relation between peak current and voltage:

E0
𝑖0 =
𝑋𝐶
Relation between RMS current and voltage:
E𝑟𝑚𝑠
𝑖𝑟𝑚𝑠 =
𝑋𝐶

Instantaneous voltage:
E = E0 sin 𝜔𝑡
Instantaneous current:
𝜋
𝑖 = 𝑖0 cos 𝜔𝑡 = 𝑖0 sin 𝜔𝑡 +
2
• 𝜋
In purely capacitive circuit, current leads voltage by 2 angle.
 Instantaneous and Average Power

Instantaneous power supplied to the capacitor:

𝜋
𝑃 = E ⋅ 𝑖 = E0 sin 𝜔𝑡 ⋅ 𝑖0 sin 𝜔𝑡 +
2
𝑃 = 𝑖0 E0 sin 𝜔𝑡 cos 𝜔𝑡

𝑖0 E0
𝑃= sin 2𝜔𝑡
2

Average power supplied to the capacitor:

𝑖0 E0
𝑃𝑎𝑣𝑔 =<𝑃 >= < sin 2𝜔𝑡 >
2

𝑃𝑎𝑣𝑔 = 0
 DC Voltage Applied to a Capacitor

• When a capacitor is connected to a

voltage source in a dc circuit, current
will flow for the short time required to
+ −
charge the capacitor. + −
+ −
𝐶
• As the charge accumulates on the
capacitor plates, the voltage across
them increases, opposing the current.
A
E0

• When the capacitor is fully charged,

the current in the circuit falls to zero.
 AC Voltage Applied to a Capacitor

For 𝟎 → 𝟏 For 𝟏 → 𝟐

Charging
Discharging

E ⇒ +𝑣𝑒
E ⇒ +𝑣𝑒
𝑖 ⇒ +𝑣𝑒 Energy is absorbed from 𝑖 ⇒ −𝑣𝑒 Energy is being returned to the
the source source
𝑃 ⇒ +𝑣𝑒 𝑃 ⇒ −𝑣𝑒
 AC Voltage Applied to a Capacitor

For 𝟐 → 𝟑 For 𝟑 → 𝟒

Charging Discharging

E ⇒ −𝑣𝑒 E ⇒ −𝑣𝑒

𝑖 ⇒ −𝑣𝑒 Energy is absorbed from the source 𝑖 ⇒ +𝑣𝑒 Energy is being returned to the
source
𝑃 ⇒ +𝑣𝑒 𝑃 ⇒ −𝑣𝑒
 Find the equation of current in the circuit shown.
T

Given: E0 = 20 𝑉 ; 𝜔 = 10 ; 𝜙 = 60°

2 𝜇𝐹
To Find: Equation of current

1 1 4Ω
Solution: 𝑋𝐶 = = = 5 × 10
𝜔𝐶 10 × 2 × 10−6

E0 20 −4
E = 20 sin(10𝑡 + 60°)
𝑖0 = = = 4 × 10 𝐴
𝑋𝐶 5 × 104

Equation of current : 𝑖 = 𝑖0 sin 𝜔𝑡 + 𝜙 + 90°

𝑖 = 4 × 10−4 sin 10𝑡 + 60° + 90°

𝑖 = 4 × 10−4 sin 10𝑡 + 150°

 Purely Inductive Circuit

𝑑𝑖
ℰ 𝑡 =𝐿
𝑑𝑡
𝐿
𝑖𝑜 sin(𝜔𝑡) 𝜋
ℰ 𝑡 = ℰ0 sin 𝜔𝑡 +
2

ℰ 𝑡 ℰ0 = 𝑋𝐿 𝑖0 ℰ𝑟𝑚𝑠 = 𝑋𝐿 𝑖𝑟𝑚𝑠

Inductive circuit
𝑋𝐿 = 𝜔𝐿 = 2𝜋𝜐𝐿

• 𝑋𝐿 is inductive reactance
• 𝐿 is inductance
 Purely Inductive Circuit

𝐿 𝜋
ℰ
𝑖(𝑡) 2

𝑖
ℰ 𝑡
Phasor diagram

Inductive circuit 𝑖 ( 𝑡 ) , ℰ 𝑡 versus 𝜔𝑡 graph (𝑋𝐿 > 1 Ω)

𝜋
ℰ 𝑡 = ℰ0 sin 𝜔𝑡 +
2 𝑖(𝑡) = 𝑖0 sin 𝜔𝑡 + 𝜙

• The inductive voltage leads current by 𝜋/2.

𝜋
ℰ(𝑡) = ℰ0 sin 𝜔𝑡 + 𝜙 +
2
 Purely Inductive Circuit with DC and AC Source

𝐿 𝐿
𝑖 𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖 𝑡 = 𝑖𝑜 sin(𝜔𝑡)

𝐼𝑟𝑚𝑠
A
𝜋
ℰ0 ℰ 𝑡 = ℰ0 sin 𝜔𝑡 +
2

Inductive circuit with Inductive circuit with

constant DC source AC source

• An inductor acts like a conducting wire in a DC circuit after some time.

 AC Voltage Applied to an Inductor

For 𝟎 → 𝟏 For 𝟏 → 𝟐

Charging Discharging
E ⇒ +𝑣𝑒
E ⇒ −𝑣𝑒
𝑖 ⇒ +𝑣𝑒 Energy is absorbed
from the source 𝑖 ⇒ +𝑣𝑒 Energy is being
𝑃 = E ⋅ 𝑖 ⇒ +𝑣𝑒 returned to the source
𝑃 = E ⋅ 𝑖 ⇒ −𝑣𝑒
 AC Voltage Applied to an Inductor

For 𝟐 → 𝟑 For 𝟑 → 𝟒

Charging Discharging

E ⇒ −𝑣𝑒 E ⇒ +𝑣𝑒
Energy is absorbed
𝑖 ⇒ −𝑣𝑒 from the source 𝑖 ⇒ −𝑣𝑒 Energy is being returned to
the source
𝑃 = E ⋅ 𝑖 ⇒ +𝑣𝑒 𝑃 = E ⋅ 𝑖 ⇒ −𝑣𝑒
 Impedance (𝒁)

• The opposition offered by an electric

circuit to an alternating current by the
combined effect of resistance and
reactance is known as impedance.

• 𝑋 is reactance which is equal to

impedance(𝑍) in purely inductive
or capacitive circuit.

ℰ0 ℰ𝑟𝑚𝑠
𝑍= =
𝑖0 𝑖𝑟𝑚𝑠
 LR Circuit

ℰ 𝑡 = ℰ0 sin(𝜔𝑡 + 𝛼) ℰ0 = 𝑖0 𝑍 = 𝑖0 𝑅 2 + X 𝐿2

ℰ 𝑡 = ℰ𝑜 sin(𝜔𝑡 + 𝜙 + 𝛼) ℰ0 = 𝑖0 𝑍 = 𝑖0 𝑅 2 + X 𝐿2
 RC Circuit

𝑉𝑅 𝑉𝐶
𝑅 𝑋𝐶
𝛼 = −tan−1
𝛼 𝑅
𝑅 𝑋𝐶
1
𝑋𝐶 = 𝑖(𝑡) = 𝑖𝑜 sin(𝜔𝑡 + 𝜙)
𝜔𝐶
𝑍= 𝑅2 + 𝑋𝐶2

ℰ 𝑡 = ℰ 0 sin(𝜔𝑡 + 𝜙 + 𝛼)
ℰ(𝑡) = 𝑉𝑅 (𝑡) + 𝑉𝐶 (𝑡)
𝜋
ℰ(𝑡) = 𝑅𝑖𝑜 sin(𝜔𝑡 + 𝜙) + 𝑋𝐶 𝑖𝑜 sin 𝜔𝑡 + 𝜙 −
2
𝜋
ℰ(𝑡) = 𝑖𝑜 𝑅 sin 𝜔𝑡 + 𝜙 + 𝑋𝐶 sin 𝜔𝑡 + 𝜙 −
2
𝑋𝐶
ℰ(𝑡) = 𝑖𝑜 𝑍 sin 𝜔𝑡 + 𝜙 − tan−1
𝑅
 RC Circuit

𝑉𝑅 𝑉𝐶
𝑋𝐶
𝑖(𝑡) 𝛼= −tan−1
𝑅
𝛼 𝑅 𝑋𝐶
ℰ(𝑡) 𝑖(𝑡) = 𝑖𝑜 sin(𝜔𝑡 + 𝜙)

ℰ 𝑡 = ℰ 0 sin(𝜔𝑡 + 𝜙 + 𝛼)

𝑋𝐶
ℰ 𝑡 = ℰ𝑜 sin 𝜔t + 𝜙 − tan−1
𝑅

ℰ𝑜 = 𝑖0 𝑍 = 𝑖0 𝑅2 + X 𝐶2
 Find 𝑉𝑜 , 𝑍, 𝑖𝑜 , 𝑖𝑟𝑚𝑠 , 𝑉𝑟𝑚𝑠 , 𝑖 𝑡 , 𝑉𝑅 𝑡 and 𝑉𝐶 (𝑡) for the given circuit.

Solution:

𝑉0 = ℰ0 = 20 V 𝑅 =3Ω 𝑋𝐶 = 4 Ω
3
𝑍= 𝑅2 + X 𝐶2 = 5 Ω

𝛼 𝛼 = 53°
4
𝑉 𝑡 = 20 sin(𝜔𝑡 + 60°)

ℰ0 20
ℰ0 = 𝑍𝑖0 ⇒ 𝑖0 = = =4A
𝑍 5
𝑖0 4 𝑉0 20
𝑖𝑟𝑚𝑠 = = =2 2A 𝑉𝑟𝑚𝑠 = = = 10 2 A
2 2 2 2
 Find 𝑉𝑜 , 𝑍, 𝑖𝑜 , 𝑖𝑟𝑚𝑠 , 𝑉𝑟𝑚𝑠 , 𝑖 𝑡 , 𝑉𝑅 𝑡 and 𝑉𝐶 (𝑡) for the given circuit.

Solution:
𝑖(𝑡) = 4 sin(𝜔𝑡 + 113°)
𝑅 =3Ω 𝑋𝐶 = 4 Ω
𝛼 = 53°

𝑉 𝑡 = 20 sin(𝜔𝑡 + 60°)

𝑉𝑅 𝑡 = 𝑖0 𝑅 sin(𝜔𝑡 + 𝜃)

𝑉𝑅 𝑡 = 12 sin(𝜔𝑡 + 113°)
𝜋
𝑉𝐶 𝑡 = 𝑖0 𝑋𝐶 sin 𝜔𝑡 + 𝜃 −
2
𝑉𝐶 𝑡 = 16 sin(𝜔𝑡 + 23°)
 Power

If the instantaneous current and voltage

in a given circuit are,
𝑖 = 𝑖 0 sin(𝜔𝑡)
𝑉 = 𝑉0 sin(𝜔𝑡 + 𝜃)
Then power is,
𝑉
𝑃(𝑡) = 𝑖 𝑡 𝑉(𝑡)
𝜃
𝑖 𝑇
‫׬‬0 𝑃 𝑡 𝑑 𝑡
𝑃𝑎 𝑣 = 𝑇
‫׬‬0 𝑑 𝑡

𝐸 = 𝑃𝑎 𝑣 𝑡

𝑇
‫׬‬0 𝑃 𝑡 𝑑 𝑡 𝑖 0 𝑉0 c o s 𝜃
𝑃𝑎 𝑣 = 𝑇 =
‫׬‬0 𝑑 𝑡 2
 Power Factor

𝑖 0 𝑉0 c o s 𝜃 𝑖0 𝑉0
𝑃𝑎 𝑣 = = c o s 𝜃 = 𝑖 𝑟 𝑚 𝑠 𝑉𝑟 𝑚 𝑠 c o s 𝜃
2 2 2

𝑃𝑎 𝑣 𝑔 = 𝑖 𝑟2𝑚 𝑠 𝑍 c o s 𝜃 𝑃𝑎 𝑣 𝑔 = 𝑖 𝑟2𝑚 𝑠 𝑅
𝑅
cos 𝜃 =
𝑍
𝑉𝑟2𝑚 𝑠 𝑉𝑟2𝑚 𝑠
𝑃𝑎 𝑣 𝑔 = cos 𝜃 𝑃𝑎 𝑣 𝑔 = 𝑅
𝑍 𝑍2

• Active Power: Portion of power absorbed by the load 𝑃𝑎 𝑐 𝑡 = 𝑉𝑟 𝑚 𝑠 𝑖 𝑟 𝑚 𝑠 c o s 𝜃

• Apparent Power: Product of rms voltage and rms current 𝑃𝑎 𝑝 𝑝 = 𝑉𝑟 𝑚 𝑠 𝑖 𝑟 𝑚 𝑠

𝑃𝑎 𝑐 𝑡
• Power Factor: Ratio of active power to apparent power
𝑃𝑎 𝑝 𝑝
= cos 𝜃
 Power Factor in Pure Circuit

Impedance
Circuit Element 𝒁 𝜃 cos 𝜽
Diagram

Purely
Resistor 𝑅 𝑅 0° 1
resistive

Purely 1
Capacitor 𝑋𝐶 90° 0
capacitive 𝜔𝐶

Purely 𝑋𝐿
Inductor 𝜔𝐿 −90° 0
inductive
 Power Factor in LR and RC Circuits

Impedance
Circuit Elements 𝒁 𝜽 cos 𝜽
Diagram

𝒁
Resistor 𝑋𝐿 𝑋𝐿
LR Circuit + 𝑅2 + X 𝐿2 𝜽 tan −1 𝑅/𝑍
𝑅
Inductor
𝑅

𝑅
Capacitor 𝜽 𝑋𝐶 𝑅/𝑍
RC Circuit + 𝑅2 + X 𝐶2 𝑋𝐶 tan −1
𝑅
Resistor
𝒁
 If a 20 𝑉 DC supply is connected to an inductor, then 5 𝐴 current runs through the
inductor after a long time. Whereas if a 20 𝑉 AC supply at 50 𝐻𝑧 is applied to the
inductor, then 4 𝐴 current runs through it. Find 𝑅, 𝐿 and the power factor.

𝑅 𝑅 𝐿
𝐿
𝑖 =5𝐴 𝑖𝑟𝑚𝑠 = 4 𝐴
20 𝑉 𝑉𝑟𝑚𝑠 = 20 𝑉
+ −

𝑉𝑟𝑚𝑠 20 𝑅
𝑉 = 20 𝑉, 𝑖 = 5 𝐴 𝑍= = ⇒𝑍 =5Ω cos 𝜃 =
𝑖𝑟𝑚𝑠 4 𝑍

𝑉
𝑅= cos 𝜃 = 0.8
𝑖 𝑍= 𝑅2 + 𝑋𝐿2 ⇒ 𝑋𝐿 = 3 Ω

𝑅 =4Ω 3
𝐿= 𝐻
100𝜋
 Choke Coil

• At most places, the household electric power

is supplied at 220 𝑉, 50 𝐻𝑧.
• If such a source is directly connected to a
mercury tube, the tube will be damaged.

• We can add resistance in series with tube

to reduce the current.
• But the power loss will be high across the
resistance box due to its high resistance.

• Choke coil is simply a coil having a large

inductance but a small resistance.
• To reduce the current choke coil is connected
in series.
 Choke Coil

• Representing the tube by a resistor and

the choke coil by an ideal inductor.
• Impedance: 𝑍 = 𝑅2 + (𝜔𝐿)2
𝑉0
• Peak current: 𝑖0 = 𝑅 𝐿
𝑅2 + (𝜔𝐿)2
𝑉𝑟𝑚𝑠
• RMS current: 𝑖𝑟𝑚𝑠 =
𝑅2 + (𝜔𝐿)2
• The rms voltage across the resistor is
𝑅 𝑉 = 𝑉0 sin(𝜔𝑡)
𝑉𝑅,𝑟𝑚𝑠 = 𝑅𝑖𝑟𝑚𝑠 = 𝑉𝑟𝑚𝑠
2
𝑅 + (𝜔𝐿) 2
• By using choke coil , the voltage across the
resistor is reduced by a factor
𝑅
𝑅2 + (𝜔𝐿)2
• The advantage of using a choke coil to
reduce the voltage is that an inductor does
not consume power.
 LC Circuit
• Current in the circuit:
𝑖 = 𝑖0 sin(𝜔𝑡)
• Voltage across capacitor:
𝑋𝐶 𝑋𝐿 𝜋
𝑉𝐶 𝑡 = 𝑖0 𝑋𝐶 sin 𝜔𝑡 −
2
• Voltage across inductor:
𝜋
𝑖 = 𝑖0 sin(𝜔𝑡) 𝑉𝐿 𝑡 = 𝑖0 𝑋𝐿 sin 𝜔𝑡 +
2
• Net voltage in the circuit:
𝑉 𝑡 = 𝑉𝐶 𝑡 + 𝑉𝐿 𝑡

𝑉(𝑡) 𝜋
𝑉 𝑡 = 𝑖 0 ( 𝑋𝐿 − 𝑋𝐶 ) s i n ( 𝜔 𝑡 + )
2
• Impedance of the circuit :
1
𝑍 = 𝑋𝐿 − 𝑋𝐶 = 𝜔𝐿 −
𝜔𝐶
 LC Circuit

Case 1: 𝑋𝐿 > 𝑋𝐶

𝜋
𝑉 𝑡 = 𝑖0𝑍 sin(𝜔𝑡 + )
2

𝑍 = 𝑋𝐿 − 𝑋𝐶
Voltage leads current

Case 2: 𝑋𝐶 > 𝑋𝐿

𝜋
𝑉 𝑡 = −𝑖0𝑍 sin(𝜔𝑡 + )
2
𝑍 = 𝑋𝐿 − 𝑋𝐶
𝜋
𝑉 𝑡 = 𝑖0𝑍 sin(𝜔𝑡 + − 𝜋)
2
𝜋 Current leads voltage
𝑉 𝑡 = 𝑖0𝑍 sin(𝜔𝑡 − )
2
 LC Circuit

𝑋𝐶 𝑋𝐿

• Current flowing through 𝐿𝐶 circuit is called

wattless current.
𝑖
𝑉(𝑡) Power factor = 0
 LCR Circuit

𝑋𝐿 𝑋𝐶 𝑅
𝑉

𝑉𝐿
𝑖 = 𝑖0 sin(𝜔𝑡)
𝜃
𝑉𝐿 − 𝑉𝐶 𝑖
𝜔𝑡

𝑉(𝑡)

𝜋
𝑉𝐶 𝑡 = 𝑖0 𝑋𝐶 sin 𝜔𝑡 − 𝑉𝐶
2
𝜋
𝑉𝐿 𝑡 = 𝑖0 𝑋𝐿 sin 𝜔𝑡 +
2

𝑉𝑅 𝑡 = 𝑖0 𝑅 sin 𝜔𝑡

𝑉 𝑡 = 𝑖0𝑍 sin(𝜔𝑡 + 𝜃) 𝑍= 𝑅 2 + 𝑋𝐿 − 𝑋𝐶 2
 LCR Circuit

Case 1: 𝑋𝐿 > 𝑋𝐶 𝑋𝐿 − 𝑋𝐶 𝑉
𝑍

𝑋𝐿 − 𝑋𝐶 𝜃
𝜃 = t a n −1 𝜃
𝑅
𝑅 𝑖

𝑅 𝑖
𝜃
Case 2: 𝑋𝐶 > 𝑋𝐿
𝜃

𝑍
𝑋𝐶 − 𝑋𝐿
𝜃 = t a n −1 𝑋𝐶 − 𝑋𝐿 𝑉
𝑅
 Resonant Frequency

A series 𝐿𝐶𝑅 circuit is said to be in the resonance condition when the current
through it has the maximum value.
• Current:
𝑋𝐿 𝑋𝐶 𝑅
𝑉0
𝑖 = s i n( 𝜔 𝑡 )
𝑍
• Current will be maximum when 𝑍 is
minimum.
𝛧 = 𝑋𝐿 − 𝑋𝐶 2 + 𝑅 2 𝑖 = 𝑖0 sin(𝜔𝑡)
𝑋𝐿 = 𝑋𝐶
1
𝜔 =
𝐿𝐶 𝑉 = 𝑉0 sin(𝜔𝑡)

1
𝑍 = 𝑅 𝑓0 =
2𝜋 𝐿𝐶

Resonant
frequency
 Resonant Frequency

𝛧 2
𝛧 = 𝑋𝐿 − 𝑋𝐶 + 𝑅2

2
1
𝛧 = 2𝜋𝑓𝐿 − + 𝑅2
2𝜋𝑓𝐶

𝛧𝑚𝑖𝑛

𝑍𝑚𝑖𝑛 = 𝑅
𝑓 < 𝑓0 𝑓0 𝑓 > 𝑓0 𝑓
 Resonant Frequency

Current is maximum Impedance is minimum

LCR circuit 𝑉𝑟𝑚𝑠 of source is 𝑉𝑟𝑚𝑠

Voltage and Current
are in phase at across resistance
resonance

Length of 𝑉𝐿 and
Power factor = 0 𝑉𝐶 phasor is same
 Resonant Frequency

𝑋𝐿 , 𝑋𝐶 , 𝑅

Resistance is
1 𝑋𝐿 independent of
𝑋𝐶 = frequency.
𝜔𝐶
𝑅

𝑋𝐶
𝑋𝐿 = 𝜔𝐿

𝜔0 𝜔
 Radio Tuning

To hear any radio station, frequency tuning is required.

1 Resonant Frequencies
𝑓0 = 𝑖 (𝐴)
2𝜋 𝐿𝐶

• Rotating the knob changes

Knob
𝐶 or 𝐿 due to which
frequency changes.

• When the frequency of the

radio matches the
broadcasted frequency, the
current is maximum. 𝑓(𝐻𝑧)
𝑂
(91.9) 𝑓0 (92.7) (98.3)
• Hence, we can tune into the
Radio Mango Big FM Radio Mirchi
FM channel.
 Sharpness of Resonance

• Resonance curve
𝑖𝑜
𝑃𝑚𝑎𝑥 = (𝑖 0 𝑚 𝑎 𝑥 )2 𝑅

𝑖 0 𝑚𝑎𝑥 2
𝑖 0𝑚𝑎𝑥 𝑃𝑚𝑎𝑥
𝑃= 𝑅=
2 2
𝑖 0𝑚𝑎𝑥
2
𝜔1 = 𝜔𝑜 + ∆𝜔
∆𝜔 ∆𝜔 𝜔2 = 𝜔𝑜 − ∆𝜔
𝑅

𝜔2 𝜔𝑜 𝜔1 𝜔
 Sharpness of Resonance

• Band width
𝜔1 − 𝜔2 = 2∆𝜔
• Measure of sharpness of Resonance 𝜔1 = 𝜔𝑜 + ∆𝜔
𝑖𝑜
𝜔0
𝜔2 = 𝜔𝑜 − ∆𝜔
2∆𝜔
𝑖 0𝑚𝑎𝑥
𝑖 0𝑚𝑎𝑥
𝑖0 = Band width
2 𝑖 0𝑚𝑎𝑥
𝑉0 𝑉0 2 𝑅
⇒ =
2 2𝑅
1 2 ∆𝜔 ∆𝜔
𝜔𝐶 − 𝜔𝐿 + 𝑅 2
2
1
⇒ 2𝑅 = − 𝜔𝐿 + 𝑅2
𝜔𝐶
2 𝜔2 𝜔𝑜 𝜔1 𝜔
2
1
⇒𝑅 = − 𝜔𝐿
𝜔𝐶
1
𝑅= − 𝜔𝐿
𝜔𝐶
 Sharpness of Resonance

• At 𝜔 = 𝜔1

1
𝑅= − 𝜔1 𝐿 𝜔1 = 𝜔𝑜 + ∆𝜔
𝜔1 𝐶 𝑖𝑜
𝜔2 = 𝜔𝑜 − ∆𝜔
1 𝑖 0𝑚𝑎𝑥
⇒𝑅= − (𝜔𝑜 + ∆𝜔)𝐿
𝐶(𝜔𝑜 + ∆𝜔) Band width
𝑖 0𝑚𝑎𝑥
2 𝑅
1 ∆𝜔
⇒𝑅= − 𝜔𝑜 𝐿 1 +
∆𝜔 𝜔𝑜 ∆𝜔 ∆𝜔
𝐶𝜔𝑜 1+ 𝜔
𝑜

∆𝜔 ∆𝜔 𝜔2 𝜔𝑜 𝜔1 𝜔
⇒ 𝑅 = 𝜔𝑜 𝐿 1 − − 𝜔𝑜 𝐿 1 +
𝜔𝑜 𝜔𝑜
1
= 𝜔0 𝐿, ∆𝜔 ≪ 𝜔0
𝐶𝜔𝑜
 Sharpness of Resonance

• At 𝜔 = 𝜔1

∆𝜔 ∆𝜔 𝜔1 = 𝜔𝑜 + ∆𝜔
⇒ 𝑅 = 𝜔𝑜 𝐿 1 − − 𝜔𝑜 𝐿 1 + 𝑖𝑜
𝜔𝑜 𝜔𝑜
𝜔2 = 𝜔𝑜 − ∆𝜔
𝑖 0𝑚𝑎𝑥
2∆𝜔
⇒ 𝑅 = 𝜔𝑜 𝐿 − Band width
𝜔𝑜 𝑖 0𝑚𝑎𝑥
2 𝑅
• Measure of sharpness of Resonance
∆𝜔 ∆𝜔
𝜔𝑜 𝜔𝑜 𝐿 1 𝐿
= = = 𝑄 Quality factor
2∆𝜔 𝑅 𝑅 𝐶
𝜔2 𝜔𝑜 𝜔1 𝜔

𝑅
∆𝜔 =
2𝐿
 Sharpness of Resonance

𝜔0 ω0 𝐿
= =𝑄 𝑖0
2∆𝜔 𝑅
𝑄1 𝑅1
𝑅1 < 𝑅2 𝑖 0𝑚𝑎𝑥 2∆𝜔 1

2
𝜔0 𝜔0 2∆𝜔
= 𝑄1 > = 𝑄2 2
2∆𝜔 1 2∆𝜔 2 𝑖 0𝑚𝑎𝑥
2
𝑄2 𝑅2
Higher the quality factor(𝑄), lesser the band
width (2∆𝜔) and sharper the resonance.
𝜔𝑜 𝜔
In a series resonant 𝐿𝐶𝑅 circuit, the voltage across 𝑅 is 100 𝑉 and 𝑅 = 1 𝑘Ω with 𝐶 =
2 𝜇𝐹. The resonant frequency 𝜔 is 200 𝑟𝑎𝑑/𝑠. At resonance the voltage across 𝐿 is

Solution:

𝑍= 𝑅 2 + 𝑋𝐿 − 𝑋𝐶 2 At resonance, 𝑋𝐿 = 𝑋𝐶 ⇒ 𝑍 = 𝑅 = 1000 Ω

𝑉= 𝑉𝑅2 + 𝑉𝐿 − 𝑉𝐶 2 At resonance, 𝑉𝐿 = 𝑉𝐶 ⇒ 𝑉 = 𝑉𝑅 = 100 𝑉

𝑉𝑅 100
∴ 𝑖= = = 0.1 𝐴
𝑅 1000
𝑖 0.1
∴ 𝑉𝐿 = 𝑖𝑋𝐿 = 𝑖𝑋𝐶 = =
𝜔𝐶 200 × 2 × 10−6

𝑉𝐿 = 250 𝑉
 Energy in LC Oscillations

−𝑞 𝑞
Total energy:

𝐶 𝑞02 (Initially, when charge is max,

𝑖 𝐸=
2𝐶 current is zero)
𝑞 2 𝐿𝑖 2
= + (At any general time)
2𝐶 2
𝐿 2
𝐿𝑖𝑚𝑎𝑥 (When current is max and
=
2 charge on capacitor is zero)
Electric
Energy
Magnetic
Energy
 LC Oscillations

• Angular frequency
1
−𝑞 𝑞 𝜔 = 2𝜋𝑓 =
𝐿𝐶

• Charge on capacitor
𝐶 𝑖 𝑞 = 𝑞0 cos 𝜔𝑡

• Current through inductor

𝑑𝑞
𝐿 𝑖=− = 𝑞0 𝜔 𝑠𝑖𝑛 𝜔𝑡
𝑑𝑡
𝑖 = 𝑖𝑚𝑎𝑥 𝑠𝑖𝑛 𝜔𝑡

• When 𝑞 is maximum, 𝑖 is zero, and vice

versa
An 𝐿𝐶 circuit like the one in the figure below contains an 80 𝑚𝐻 inductor and a
5 𝑚𝐹 capacitor that initially carries a 185 𝜇𝐶. The switch is open for 𝑡 < 0 and is
then thrown closed at 𝑡 = 0.
a) Find the frequency (in 𝐻𝑧) of the resulting oscillations.
b) Find the maximum current in the circuit.

Solution:
1 1
a) 𝑓 = =
2𝜋 𝐿𝐶 2𝜋 80 × 10−3 × 5 × 10−3
+ 𝐿
𝐶
𝑓 = 7.96 𝐻𝑧
−

b) Maximum current in the circuit, 𝑖0 = 𝑞0 𝜔

185 × 10−6 𝑆
=
80 × 10−3 × 5 × 10−3

𝑖0 = 9.25 𝑚𝐴
 Parallel AC Circuits

Series Circuits Parallel Circuits

• Current phasor is taken as reference • Voltage phasor is taken as reference

• Voltage phasors are added for each • Current phasors are added for each
element branch
(however, individual branches need
to be solved separately using series)

𝑋𝐿 𝑅
𝑋𝐿 𝑋𝐶
𝑅

𝑖2 𝑋𝐶 𝑅
𝑖1

E𝑜 sin 𝜔𝑡
Voltage relation in transformer:

ε𝑃 𝑁𝑃
=
ε𝑆 𝑁𝑆

𝑖𝑃 𝑖𝑆
Power relation in transformer: 𝜀𝑆
𝜀𝑃 𝑁𝑃 𝑁𝑆

𝑖𝑆 ε𝑃 𝑁𝑃 For ideal
= =
𝑖𝑃 ε𝑆 𝑁𝑆 transformer

Efficiency of transformer:

𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 ε𝑆 ∙ 𝑖𝑆
𝜂= =
𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 ε𝑃 ∙ 𝑖𝑃
 Efficiency of Transformer

Laminated Core
Primary winding
• Input power
= Output power + Losses

𝑖𝑃 𝑖𝑆 • Efficiency
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
E𝑆 𝜂=
E𝑃 𝑁𝑃 𝑁𝑆 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
𝜂% = × 100
𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟

AC Output • Real transformers have efficiency

supply greater than 90%

Secondary winding
 Step-up Transformer Step-down Transformer

𝑵𝑷 < 𝑵𝑺 𝑵𝑷 > 𝑵𝑺

Using the formula Using the formula

𝑖𝑆 E𝑃 𝑁𝑃 𝑖𝑆 E𝑃 𝑁𝑃
= = = =
𝑖𝑃 E𝑆 𝑁𝑆 𝑖𝑃 E𝑆 𝑁𝑆
E𝑃 < E𝑆 Voltage Increases E𝑃 > E𝑆 Voltage Decreases
𝒊𝑷 > 𝒊𝑺 Current Decreases 𝒊𝑷 < 𝒊𝑺 Current Increases
 Losses in Transformer

Winding losses Eddy current

losses

𝑖𝑃 𝑖𝑆

E𝑃 𝑁𝑃 𝑁𝑆 E𝑆

AC Output
supply

Flux losses Hysteresis

losses
Winding Losses
• Winding loss:
Heat = ‫ 𝑖 ׬‬2 𝑅𝑑𝑡
• To minimize them:
• Use material of low resistivity like copper
for windings
• Use thick wires as
1
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 ∝
𝐴𝑟𝑒𝑎

Flux Losses

• Flux loss:
• All flux from primary winding does not
get linked to secondary winding
• Some flux leaks out of the core
• To minimize them:
• Use highly ferromagnetic core like soft
iron as it maximizes flux linkage
Hysteresis Losses
• Hysteresis loss:
• Continuous magnetization-demagnetization causes
energy loss through hysteresis
• To minimize them:
• Use soft iron for transformer core as its hysteresis loop
has smaller area
Energy Loss ∝ Area of Hysteresis Loop

Eddy Losses

• Eddy loss:
• Due to alternating current, flux changes
in the core continuously
• This generates eddy currents and
hence, large amount of heat
• To minimize the loss:
• Use laminated core instead of solid core
 𝑡
The electric current in a circuit is given by 𝑖 = 𝑖0 for some time. The rms current
𝜏
T for the period 𝑡 = 0 to 𝑡 = 𝜏 is:

𝑡
Given: 𝑖 = 𝑖0
𝜏

To find: RMS current for 𝑡 = 0 to 𝑡 = 𝜏

Solution: 𝜏 𝜏 𝑡 2
‫׬‬0 𝑖 2 𝑑𝑡 ‫׬‬0 𝑖02 𝜏
𝑑𝑡
Mean square current, <𝑖 >=2
𝜏 =
‫׬‬0 𝑑𝑡 𝜏

2 𝜏 2
𝑖0 𝑖0
⇒ < 𝑖 2 > = 3 න 𝑡 2 𝑑𝑡 =
𝜏 0 3

RMS current, 𝑖𝑟𝑚𝑠 = < 𝑖2 >

𝑖0
𝑖𝑟𝑚𝑠 =
3
 An alternating voltage 𝑉 𝑡 = 220 sin 100𝜋𝑡 volt is applied to a purely resistive load of
5 Ω. The time taken for the current to rise from half of the peak value to the peak
T value is

Given: 𝑉 𝑡 = 220 sin 100𝜋𝑡 , 𝑅 = 5 Ω

𝑖0
To find: Time taken for the current rise from to 𝑖0
2
220
Solution: 𝑖 𝑡 =
5
sin 100𝜋𝑡

⇒ 𝑖 𝑡 = 44 sin 100𝜋𝑡

𝑖0 10
For 𝑖 𝑡 = = 22 𝐴: 22 = 44 sin 100𝜋𝑡1 ⇒ 𝑡1 = 𝑚𝑠
2 6
For 𝑖 𝑡 = 𝑖0 = 44 𝐴: 44 = 44 sin 100𝜋𝑡2 ⇒ 𝑡2 = 5 𝑚𝑠

𝑖0 10
Time taken for the current rise from 2
to 𝑖0 = 𝑡2 − 𝑡1 = 5 − = 3.3 𝑚𝑠
6
 In series LR circuit, 𝑋𝐿 = 3𝑅. Now a capacitor with 𝑋𝐶 = 𝑅 is connected in series.
The ratio of new to old power factor is:

Given: 𝑋𝐿 = 3𝑅, 𝑋𝐶 = 𝑅
cos 𝜙𝑓 𝑅 𝑋𝐿 = 3𝑅
To find: Ratio
cos 𝜙𝑖
Solution:
For LR circuit: For LCR circuit:

𝑍𝑖 = 𝑅2 + 3𝑅 2 𝑍𝑓 = 𝑅2 + 3𝑅 − 𝑅 2

⇒ 𝑍𝑖 = 𝑅 10 ⇒ 𝑍𝑓 = 𝑅 5
𝑅 𝑋𝐿 = 3𝑅 𝑋𝐶 = R
𝑅 1 𝑅 1
cos 𝜙𝑖 = = cos 𝜙𝑓 = =
𝑍𝑖 10 𝑍𝑓 5

cos 𝜙𝑓
Ratio = = 2
cos 𝜙𝑖