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What is the power dissipation (in W) in an ideal inductor having an inductance of 0.2 H?
AC through Pure Inductor:
Consider a pure inductor of inductance L connected to an alternating voltage given by,
v = Vm sin ωt
An alternating current (i) flows through the inductor and sets up a changing magnetic field and it induced a back EMF (L di/dt).
By using KVL,
Integrating on both side,
The phasor diagram can be drawn as,
From the phasor diagram, it is clear that the phase angle between voltage and current is 90°.
We know that the power factor is the cosine angle of voltage and current,
Hence, Power Factor = cos ϕ = cos 90° = 0
We know that,
Power Dissipation (P) = VIcos ϕ
Since, ϕ = 90° ⇒ cos ϕ = cos 90° = 0
Hence, Power Dissipation (P) = 0
Consider the image given below. According to the image, find out the value of L_{eq} if L is 22 mH.
Note: Calculation for series & parallel connection of inductance is same as the resistance:
Application:
We have,
It can be drawn as,
After solving it,
L_{eq} = 16 mH
Self induction:
Two coils in differential connection have a selfinductance of 2 mH and 4 mH and mutual inductance of 0.15 mH. The equivalent inductance of the series opposing combination will be:
Concept:
Series Aiding:
The equivalent inductance of series aiding connection is:
Leq = L1 + L2 + 2M
Series Opposing:
The equivalent inductance of series opposing connection is:
Leq = L1 + L2 – 2M
Calculation:
Given,
L_{1} = 2 mH
L_{2} = 4 mH
M = 0.15 mH
And connected with series opposing polarity.
From above concept,
L_{eq} = L_{1} + L_{2}  2M
L_{eq} = 2 + 4  2 (0.15)
L_{eq} = 5.7 mH
If both the number of turns and core length of an inductive coil are doubled, then its self inductance will be
CONCEPT:
SelfInduction
⇒
Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.
Given  l2 = 2l1
N_{2} = 2 N_{1}
Selfinductance of a solenoid is given by:
⇒
⇒
⇒
L_{2} = 2 L_{1}
The energy stored in a coil with inductance L is determined as:
Current: Rate of flow of charge is known as current. It can be expressed as
I = dq/dt
Where, q = Charge in coulomb
t = time in seconds
The energy stored by an inductor carrying a current I is given by:
Energy, E = 1/2LI^{2} Joule
Where, L = Inductance of the coil
Now energy stored by inductor can also be expressed as
Inductance of a solenoid of 2500 turns wound uniformly over a length of 0.5 m cylindrical paper tube of 4 cm diameter in air medium is around:
Given that, number of turns (N) = 2500
Diameter (d) = 4 cm
Length (l) = 0.5 m
The value of permeability in air is taken as 4π × 10^{7}.
Selfinductance of a solenoid is given by:
Two coils having selfinductance of 18 H and 2 H, respectively, are magnetically coupled and the mutual inductance between them is 3 H. Find the value of coefficient of coupling.
Concept:
Coefficient of Coupling (k):
The coefficient of coupling (k) between two coils is defined as the fraction of magnetic flux produced by the current in one coil that links the other.
Two coils have selfinductance L1 and L2, then mutual inductance M between them then Coefficient of Coupling (k) is given by
Where,
N_{1} and N_{2} is the number of turns in coil 1 and coil 2 respectively
A is the crosssection area
l is the length
Calculation:
Given,
L_{1} = 18 H
L_{2} = 2 H
M = 3 H
From the above concept,
If the number of turns in a coil is doubled in the same length and area, the inductance L will
Concept:
The inductance of a coil is given by:
N = number of turns in the coil
A = Area of the crosssection
l = length of the coil
μ = magnetic permeability of the coil
Application:
With N_{2} = 2 × N_{1}, A_{2} = A_{1} and l_{2} = l_{1}
∴ If the number of turns in a coil is doubled in the same length and area, the inductance L will quadruple.
If a brass core of an inductor is replaced by an iron core, the inductance of coil:
The inductance of an inductor is primarily determined by four factors:
The type of core material:
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