Electromagnetic oscillation

A Coupled Oscillation of the electric field and the magnetic field

that constitute a single electromagnetic field is called
electromagnetic oscillation. Electromagnetic oscillation
propagates as electromagnetic wave consists of oscillating
electric and magnetic fields. Electric and magnetic fields
oscillate perpendicular to each other and propagate in the
direction perpendicular to both.

L-C oscillation (Free Oscillation):-

The electric current and the charge on the capacitor in the circuit
undergo electrical LC oscillations when a charged capacitor is
connected to an inductor. The electrical energy stored in the
capacitor is its initial charge which is maximum.

The inductor contains zero energy. When the switch is turned

on, the current in the circuit starts increases and the charge on
the capacitor keeps decreasing. The current induced in the
circuit produces a magnetic field in the inductor.
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The total energy at that instant in an oscillating L-C circuit is the
sum of electric and magnetic energy.
𝑖. 𝑒. 𝑈 = 𝑈𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 + 𝑈𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐

𝑞2 𝐿𝐼 2
𝑖. 𝑒. 𝑈 = + … … … (1)
2𝐶 2
Since, the circuit has no resistance which indicate, there is no
energy loss and the total energy is constant with time.
𝑑𝑈
∴ = 0 … … … (2)
𝑑𝑡
𝑑 𝑞2 𝐿𝐼 2
𝑜𝑟, ( + )= 0
𝑑𝑡 2𝐶 2
2𝑞 𝑑𝑞 2𝐿𝐼 𝑑𝐼
𝑜𝑟, + =0
2𝐶 𝑑𝑡 2 𝑑𝑡
𝑞 𝑑𝐼
𝑜𝑟, . 𝐼 + 𝐿𝐼 = 0
𝐶 𝑑𝑡
𝑞 𝑑𝐼
𝑜𝑟, +𝐿 =0
𝐶 𝑑𝑡
𝑞 𝑑 𝑑𝑞
𝑜𝑟, +𝐿 ( )=0
𝐶 𝑑𝑡 𝑑𝑡
𝑞 𝑑2𝑞
𝑜𝑟, +𝐿 2 =0
𝐶 𝑑𝑡
𝑑2𝑞 𝑞
𝑜𝑟, 2
+ = 0 … … … (3)
𝑑𝑡 𝐿𝐶
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This is the differential equation that, describes the oscillation of
resistance less L-C circuit. The solution of this equation is;
𝑞 = 𝑞0 sin(𝑤𝑡 + ∅)
Now, comparing this equation with standard differential
equation of SHM;
𝑑2𝑦
𝑖. 𝑒. 2
+ 𝑤 2𝑦 = 0
𝑑𝑡
1
∴𝑤=
√𝐿𝐶
1
∴𝑓= … … … . (4)
2𝜋√𝐿𝐶
Hence an LC circuits oscillate with constant frequency given by,
above equation.

Oscillation of Electric and magnetic energy in an LC

circuit:-
The energy stored as the electric field in the capacitor at any
time t is;
𝑞2 [𝑞0 sin(𝑤𝑡 + ∅)]2 𝑞02
𝑈𝐸 = = = 𝑠𝑖𝑛2 (𝑤𝑡 + ∅)
2𝐶 2𝐶 2𝐶
𝑞02
(𝑈𝐸 ) 𝑚𝑎𝑥 = … … … . (5)
2𝐶

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Similarly, the energy store as magnetic field in the inductor at
the same instant is,
𝐿𝐼 2 1 𝑑𝑞 2 𝐿
𝑈𝐵 = = 𝐿( ) = [𝑤𝑞0 cos(𝑤𝑡 + ∅)]2
2 2 𝑑𝑡 2
𝑞02 2 2(
𝑞02 1
𝑈𝐵 = . 𝐿. 𝑤 𝑐𝑜𝑠 𝑤𝑡 + ∅) = . 𝐿. 𝑐𝑜𝑠 2 (𝑤𝑡 + ∅)
2 2 𝐿𝐶
𝑞02
𝑈𝐵 = 𝑐𝑜𝑠 2 (𝑤𝑡 + ∅)
2𝐶
𝑞02
(𝑈𝐵 )𝑚𝑎𝑥 = … … … . (6)
2𝐶
From equation (5) and (6) we can say that maximum value of
electric and magnetic energy stored in LC circuit is equal and at
any instant the sum of electric and magnetic energy is equal to
𝑞02
a constant.
2𝐶

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L-C oscillation with resistance (Damped oscillation):-

Consider a circuit consists of an inductor with inductance ‘L’, a

capacitance with capacitor ‘C’ and resistor of resistance ‘R’ as
shown in figure. Consider a capacitor is given a charge Qo
initially, due to the presence of resistance a part of energy is
converted in to a thermal energy. Because of this loss of energy
the oscillation of charge, current and potential difference
continuously decreases in amplitude. Such oscillations are called
damped oscillation.
Now, Using Kirchhoff’s voltage law;
𝑑𝐼 𝑄
𝐿 + 𝐼𝑅 + = 0
𝑑𝑡 𝐶
𝑑 𝑑𝑄 𝑑𝑄 𝑄
𝑜𝑟, 𝐿 ( )+ 𝑅 + =0
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐶
𝑑2𝑄 𝑑𝑄 𝑄
𝑜𝑟, 𝐿 2+ 𝑅 + = 0 … … … (1)
𝑑𝑡 𝑑𝑡 𝐶
This is the differential equation for damped LCR oscillation.

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Comparing this equation with differential equation for
mechanical damped harmonic oscillator equation;
𝑑2𝑥 𝑑𝑥
𝑜𝑟, 𝑚 2+ 𝑏 + 𝑘𝑥 = 0 … … … (2)
𝑑𝑡 𝑑𝑡
We get;
𝑚 = 𝐿, 𝑏 = 𝑅, 𝐾 = 1/𝐶
Therefore, the angular frequency of damped oscillation will be

𝐾 𝑏2 1 𝑅2
𝑤= √ − 2
= √ − 2 … … … (3)
𝑚 4𝑚 𝐿𝐶 4𝐿

The solution of equation (2) is

𝑥 = 𝐴𝑒 −𝑏𝑡/2𝑚 sin(𝑤𝑡 + ∅)
Therefore the solution of equation (1) is given by
𝑄 = 𝑄0 𝑒 −𝑅𝑡/2𝐿 sin(𝑤𝑡 + ∅) … … … (4)
Therefore, oscillation of damped (LCR) oscillation is a
sinusoidal oscillation with an exponentially decaying amplitude
𝑄0 𝑒 −𝑅𝑡/2𝐿 .
Equation (3) shows that the angular frequency of the damped
oscillation is always less than the angular frequency of un-
damped oscillation.

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From equation (3),
1 𝑅2
1. When > 4𝐿2 , 𝑤 is positive, the discharge of capacitor
𝐿𝐶
is oscillatory (under damped).
1 𝑅2
2. When = , 𝑤 is zero, the discharge is non-
𝐿𝐶 4𝐿2
oscillatory (critically damped).
1 𝑅2
3. When < , 𝑤 is negative, the discharge is non-
𝐿𝐶 4𝐿2
oscillatory (over damped).

Forced Electromagnetic oscillation:-

Since, there is always some resistance present in an electric

circuit. So some energy is converted in to heat energy causing
damped oscillation. Therefore, the periodic input of power is
given to bring it to behave as an oscillatory circuit. Such
oscillations are called forced electromagnetic oscillation. Figure

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sows LCR series circuit with an AC frequency generator of
𝑒𝑚𝑓.
𝑉 = 𝑉0 𝑠𝑖𝑛𝑤𝑡
Now, Using KVL;
𝑑𝐼 𝑄
𝐿 + 𝐼𝑅 + = 𝑉0 𝑠𝑖𝑛𝑤𝑡
𝑑𝑡 𝐶
𝑑 𝑑𝑄 𝑑𝑄 𝑄
𝑜𝑟, 𝐿 ( )+𝑅 + = 𝑉0 𝑠𝑖𝑛𝑤𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐶
𝑑2𝑄 𝑑𝑄 𝑄
𝑜𝑟, 𝐿 2 +𝑅 + = 𝑉0 𝑠𝑖𝑛𝑤𝑡 … … … (1)
𝑑𝑡 𝑑𝑡 𝐶
This is the differential equation for forced LCR oscillation.
Comparing this equation with differential equation of force
mechanical oscillation;
𝑑2𝑥 𝑑𝑥
𝑜𝑟, 𝑚 2 +𝑏 + 𝐾𝑥 = 𝐹0 𝑠𝑖𝑛𝑤𝑡 … … … (2)𝑤𝑒 𝑔𝑒𝑡,
𝑑𝑡 𝑑𝑡
1
𝑚 = 𝐿, 𝑏 = 𝑅, 𝐾= , 𝐹0 = 𝑉0
𝐶
The solution of equation (2) is;
𝑥 = 𝐴 sin(𝑤𝑡 + ∅)
Then, solution of equation (1) becomes;
𝑄 = 𝑄0 sin(𝑤𝑡 + ∅) … … … (3)

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 𝐹0 /𝑚
Since, 𝐴 =
√(𝑤0 2 − 𝑤 2 )2 + (𝑏𝑤)2
𝑚
𝑉0 /𝐿
∴ 𝑄0 =
√(𝑤0 2 − 𝑤 2 )2 + (𝑅𝑤)2
𝐿
Differentiating equation (3) w.r.t. time. Current is given by
𝑑𝑄
𝐼= = 𝑄0 𝑤 cos(𝑤𝑡 + ∅)
𝑑𝑡
This can be written as, 𝐼 = 𝐼0 cos(𝑤𝑡 + ∅)
𝑉0
Where, 𝐼0 = 𝑄0 𝑤 = … … … (4)
𝑍

Here, 𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝑐 )2 is the impedance of LCR circuit.

The inductive reactance, 𝑋𝐿 = 𝐿𝑤 = 2𝜋𝑓𝐿 and the capacitive
1 1
reactance, 𝑋𝐶 = 𝐶𝑤 = 2𝜋𝑓𝐶 .

Resonance:-

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When the applied frequency is equal to natural frequency the
amplitude of current in the circuit becomes maximum. Such
condition is called resonance. The frequency at which amplitude
of current becomes maximum is called resonance frequency.
From equation (4) it is seen that, the current in the circuit will be
maximum when Z is minimum, and Z is minimum when 𝑋𝐿 =
𝑋𝐶 .
1
⇒ 𝑤𝐿 =
𝑤𝐶
1 1
⇒𝑤= ∴ 2𝜋𝑓 =
√𝐿𝐶 √𝐿𝐶
1
Hence, Resonance frequency, 𝑓 = 2𝜋
√𝐿𝐶

1
At resonance, 𝑤 = 𝑤0 =
√𝐿𝐶
1
∴ 𝑓𝑟 =
2𝜋√𝐿𝐶
Quality Factor:-
It is defined as the ratio of the voltage drop across inductor (L)
or across capacitor (C) to the voltage drop across resistor (R) at
resonance.
𝑉𝐿 𝑜𝑟 𝑉𝑐
𝑖. 𝑒. 𝑄 =
𝑉𝑅

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It is also defined as the ratio of the energy in the inductor or
capacitor to the energy dissipated across the resistance.
𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑
𝑖. 𝑒. 𝑄 =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑
The quality factor is also defined in terms of lower and upper
half part of frequencies f1 and f2 these frequencies lying below
and above the resonance frequency. Where, the power
dissipation is half of that at resonance frequency. At these
𝐼𝑟
frequencies current in the circuit is where Ir is current at
√2
resonance frequency.
2𝜋𝑓𝑟
𝑖. 𝑒. 𝑄 =
𝑓2 − 𝑓1

Numerical Examples:-
1. A radio tuner has a frequency range from
𝟓𝟎𝟎 𝐊𝐇𝐳 𝐭𝐨 𝟓 𝐌𝐇𝐳. If its LC circuit has an effective
inductance of 𝟒𝟎𝟎 𝛍𝐇. What is the range of its variable
capacitor?
Solution:-
𝐹1 = 500 KHz = 500 × 103 𝐻𝑧
𝐹2 = 5 MHz = 5 × 106 𝐻𝑧
𝐿 = 400 𝜇𝐻 = 400 × 10−6 𝐻
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 𝐶 =?
1
We have; F1 =
2π√LC
1 1
𝑜𝑟, 𝐶 = =
4𝜋𝐿𝐹12 4𝜋 × 400 × 10−6 × (500 × 103 )2
∴ 𝐶 = 2.535 × 10−10 𝐹
1
Also; F2 =
2π√LC
1 1
𝑜𝑟, 𝐶 = =
4𝜋 2 𝐿𝐹22 4𝜋 2 × 400 × 10−6 × (5 × 106 )2
∴ 𝐶 = 2.535 × 10−12 𝐹
∴ Range of variable capacitor is 2.535 ×
10−12 𝐹 𝑡𝑜 2.535 × 10−10 𝐹.

2. A circuit has 𝐋 = 𝟏𝟎 𝐦𝐇 𝐚𝐧𝐝 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐚𝐧𝐜𝐞 𝐂 = 𝟏𝟎 𝛍𝐅.

How much resistance should be added to circuit so that
frequency of oscillation will be 𝟏 % less than that of LC
oscillation?
Solution:-
L = 10 mH = 10 × 10−3 H
C = 10 μF = 10 × 10−6 F

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 1 1
We have; F0 = =
2π√LC 2𝜋√10 × 10−3 × 10 × 10−6
= 503.29 𝐻𝑧
From question; F = F0 − 1% of F0
503.29
= 503.29 − = 498.51 𝐻𝑧
100
Then, F = F0 − 1% of F0

1 1 𝑅
𝑜𝑟, √ − ( )2 = 498.51
2𝜋 𝐿𝐶 2𝐿

1 𝑅
𝑜𝑟, √ − ( )2 = 3130.64
𝐿𝐶 2𝐿

1 𝑅
𝑜𝑟, − ( )2 = 9800924.341
𝐿𝐶 2𝐿
1
𝑜𝑟, − 9800924.341
10 × 10−3 × 10 × 10−6
𝑅2
=
4 × (10 × 10−3 )2
𝑜𝑟, 𝑅2 = 79.63
∴ 𝑅 = 8.92 𝛺

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 3. If 𝟏𝟎 𝐦𝐇 inductor and two capacitor of 𝟓 𝛍𝐅 𝐚𝐧𝐝 𝟐 𝛍𝐅
are given. Find the two resonance frequencies that can
be obtained by connecting these elements in different
ways.
Solution:-
L = 10 mH = 10 × 10−3 H
C1 = 5 μF = 5 × 10−6 F
C2 = 2 μF = 2 × 10−6 F
For series combination;
1 1 1
= +
𝐶𝑠 𝐶1 𝐶2
1 1
= + = 700000
5 × 10−6 2 × 10−6
∴ 𝐶𝑠 = 1.42 × 10−6 𝐹
1 1
𝑇ℎ𝑒𝑛; Fr = =
2π√LC 2𝜋√10 × 10−3 × 1.42 × 10−6
∴ Fr = 1335.599 Hz
For parallel combination;
𝐶𝑝 = 𝐶1 + 𝐶2 = 5 × 10−6 + 2 × 10−6 = 7 × 10−6 𝐹
1 1
Then; Fr = =
2π√LC 2𝜋√10 × 10−3 × 7 × 10−6

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 ∴ Fr = 601.55 Hz

4. In an oscillating LC circuit, what value of charge

expressed in terms of maximum charge is present on
the capacitor when the energy is shared equally
between the electric and magnetic fields? At what time
will this condition occur, assuming the capacitor is fully
charged initially? Assume that 𝑳 = 𝟏𝟎 𝒎𝑯 𝐚𝐧𝐝 𝑪 =
𝟏 𝝁𝑭.
Solution:-
𝐻𝑒𝑟𝑒, 𝐿 = 10 𝑚𝐻 = 1 × 10−3 𝐻, 𝐶 = 1𝜇𝐹 = 1 × 10−6 𝐹
According to question, 𝑈𝐸 = 𝑈𝐵
Since, 𝑈𝐸 + 𝑈𝐵 = 𝑈𝑚𝑎𝑥
2𝑈𝐸 = 𝑈𝑚𝑎𝑥
1
𝑈𝐸 = 𝑈𝑚𝑎𝑥
2
2
𝑞 1 𝑞0 2
=
2𝐶 2 2𝐶
𝑞0
∴𝑞=
√2
Also from question, at 𝑡 = 0, 𝑞 = 𝑞0 ⇒ ∅ = 0
𝑞 = 𝑞0 𝑠𝑖𝑛 𝑤𝑡
𝑞0 1 𝜋
= 𝑞0 𝑠𝑖𝑛 𝑤𝑡 ⇒ sin 𝑤𝑡 = = 𝑠𝑖𝑛
√2 √2 4

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 𝜋 𝜋 𝜋√𝐿𝐶
𝑤𝑡 = ⇒ 𝑡 = =
4 4𝑤 4
3.14 × √1 × 10−3 × 1 × 10−6
𝑡=
4
∴ 𝑡 = 2.5 × 10−4 𝑠𝑒𝑐

5. What resistance ‘R’ should be connected in series with

an inductance 220 mH and capacitance 12 µF for the
maximum charge on the capacitor to decay to 99% of its
initial value in 50 Cycles.
Solution:-
𝐻𝑒𝑟𝑒, 𝐿 = 220 × 10−3 𝐻, 𝐶 = 12 × 10−6 𝐹
2𝜋
𝑡 = 50𝑇 = 50 ( ) = 100𝜋√𝐿𝐶 = 0.5104 𝑠𝑒𝑐
𝑤0
We have, 𝑄 = 𝑄0 𝑒 −𝑅𝑡/2𝐿 sin(𝑤𝑡 + ∅)
𝑄𝑚𝑎𝑥 = 𝑄0 𝑒 −𝑅𝑡/2𝐿 , 𝑤ℎ𝑒𝑟𝑒 𝑄0 𝑖𝑠 𝑐ℎ𝑎𝑟𝑔𝑒 𝑎𝑡 𝑡 = 0𝑠𝑒𝑐

𝑅𝑡/2𝐿
𝑄0
𝑒 =
𝑄𝑚𝑎𝑥
2𝐿 𝑄0
𝑅= ln( )
𝑡 𝑄𝑚𝑎𝑥
From question, 𝑄𝑚𝑎𝑥 = 99% 𝑜𝑓 𝑄0 = 0.99𝑄0
2 × 220 × 10−3 𝑄0
∴𝑅= ln( )
0.5104 0.99𝑄0
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 = 8.66 × 10−3 𝛺

6. A series LCR circuit has inductance 12 mH, Capacitance

1.6 µF and resistance 1.5 Ω i) At what time the amplitude
of charge oscillation will be 50 % of its initial value ii) To
how many periods of oscillation does this happen.
Solution:-
𝐻𝑒𝑟𝑒, 𝐿 = 12 × 10−3 𝐻, 𝐶 = 1.6 × 10−6 𝐹, 𝑅 = 1.5𝛺
i) We have for damped LCR circuit,
𝑄 = 𝑄0 𝑒 −𝑅𝑡/2𝐿 sin(𝑤𝑡 + ∅)
∴ Amplitude of charge osillation, 𝑄𝑚 = 𝑄0 𝑒 −𝑅𝑡/2𝐿
𝑄0
From question, 𝑄𝑚 = 50%𝑜𝑓𝑄0 =
2
𝑄0
∴ = 𝑄0 𝑒 −𝑅𝑡/2𝐿 ⇒ 𝑒 𝑅𝑡/2𝐿 = 2
2
2𝐿 2 × 12 × 10−3 × ln(2)
𝑡= ln(2) =
𝑅 1.5
∴ 𝑡 = 0.011 𝑠𝑒𝑐
2𝜋
ii) Since, 𝑇 = = 2𝜋√𝐿𝐶
𝑤

= 2 × 3.14√12 × 10−3 × 1.6 × 10−6

= 8.7 × 10−4 𝑠𝑒𝑐
0.011
∴ Number of periods = = 13
8.7 × 10−4

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Exercise:-
1. Define LC oscillation qualitatively by using necessary
circuit and graph.
2. What is LC oscillation? Derive the differential equation of
free oscillation and compare its solution with mass spring
system.
3. Prove that LC circuit is an analogy of spring mass system.
Hence prove that maximum energy stored in the capacitor
is equal to maximum energy stored in inductor.
4. Define sharpness of resonance. Derive the relation for
current amplitude of forced e-m oscillation.
5. Compare the damped and forced LCR oscillation. Derive
the differential equation of forced e-m oscillation and
compare it with driven mechanical oscillation.
6. What is resonance? Formulate the differential equation of
forced e-m oscillation. Then determine the expression for
resonant frequency.
7. Derive the differential equation of the forced oscillation of
LCR circuit with an AC source and find the expression for
the current amplitude. Hence explain the condition of
current resonance in such circuit.
8. LC oscillations are called e-m oscillations, why? Derive the
differential equation for damped electromagnetic
oscillations and find the amplitude and frequency of that
oscillation.

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 9. Discuss about the damped electromagnetic oscillation. Find
the expression for damped frequency. Also discuss about
over damping, critical damping and under damping
condition.
10. Derive a relation for current flopping in the circuit
containing a resistor, an inductor and a capacitor in series
with a sinusoidal varying emf. Find the condition for
current response.
11. Obtain an expression for current in a driven LCR
circuit and discuss how the current leads or lags the applied
voltage in phase; (a) when the net reactance in circuit is
inductive and (b) when the reactance in circuit is equal to
resistance. Illustrate it with the help of a figure.
12. A 2 µF capacitor is charged up to 50 V. The battery is
disconnected and 50 mH coil is connected across the
capacitor so that LC oscillation to occur. Calculate the
maximum value of current in the circuit.
13. A circuit has L = 1.2 mH, C = 1.6 µF and R = 1.5 Ω.
(a) After what time t will the amplitude of the charge
oscillation drop to one half of its initial value. (b) To how
many periods does this correspond?
14. What should be the capacitance of a capacitor in a
tuned circuit of frequency 10 MHz having an inductance of
0.01 mH? The resistance of the circuit is negligible?
15. A 20 mH inductor and 600 µF capacitor form an
oscillating circuit. What is the peak value of current if the
initial charge is 60 µC?
Prepared by Gunanidhi Gyanwali
 16. A circuit has L = 5 mH and C = 2 µF. How much
resistance must be inserted in the circuit to reduce the
resonance frequency by 5 %.
17. Calculate the resonating frequency and quality
factor of a circuit having 0.02 µF capacitance, 8 mH
inductance and 0.25 Ω resistance.

Prepared by Gunanidhi Gyanwali