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Concept:
According to Faraday’s Law:
A pure inductor of 20.0 mH is connected to a source of 220 V. Find the RMS value of current in the circuit, if the frequency of the source is 50 Hz.
Concept:
Inductive reactance:
⇒ X_{L} = 2πfL
Where f = frequency of ac current and L = self  inductance of the coil
Impedance:
AC voltage applied to an inductor:
⇒ I = V / X_{L}
Calculation:
Given Vrms = 220 V
L = 20 mH = 20 × 103 H
f = 50 Hz
Inductive reactance is given as,
⇒ X_{L} = 2πfL
⇒ X_{L} = 2π × 50 × 20 × 103
⇒ X_{L} = 6.28 Ω
When an AC voltage is applied to an inductor, the current in the circuit is given as,
An aircored solenoid has 100 turns and is 1 m long. If the area of the cross section is 100 cm^{2}, what will be its selfinductance (in μH)?
Concept:
Selfinductance
Here, A = Area
N = Number of turns
l = Length of coil
μ_{0} = absolute permeability of material
μ_{r} = relative permeability of material
Calculation:
N = 100, l = 1 m,
A = 100 cm^{2} = 100 cm × cm = 100 × 10^{2} m × 10^{2} m = 100 × 10^{4} m^{2}
μ_{∘ }= 4 π × 10^{7}, μ_{r} = 1 (Air)
= 40 π μH
Concept:
In inductive circuit, when inductance (L) or inductive reactance (X_{L}) increases, the circuit current
Concept:
The current in an inductive circuit is given by:
I = V/X_{L}
where I = Current
V = Voltage
X_{L} = Inductive reactance
Explanation:
Key Points
When a dc voltage of 60 V is applied across a solenoid, an electric current of 5 A flows through it. When the dc voltage is replaced by the ac voltage of the angular frequency 400 rad/s, the electric current is reduced to 3A. What is the inductance of solenoid?
Concept:
Solenoid:
A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
Calculation:
Given: V_{dc} = 60 V, I_{dc} = 5 A
Solenoid equivalent is given by series RL circuit.
Now by DC supply, the inductor behaves as a short circuit at a steady state.
V_{dc} = I_{dc} × R = 5 × R = 60
R = 12 Ω
Now, when AC supply is given, Current I_{ac} = 3 A at ω = 400 rad/sec
Equivalent impedance(Z) = V/I = 60/3 = 20Ω
Therefore,
ωL = 16 Ω
L = 16/400 = 0.04 H
Two coils with selfinductance of 100 mH and 200 mH and mutual inductance of 50 mH are connected in parallel aiding order. What will be the total inductance of the combination?
Concept:
The equivalent inductance of series aiding connection is
L = L_{1} + L_{2} + 2M
The equivalent inductance of series opposing connection is
L = L_{1} + L_{2} – 2M
Equivalent inductance of parallel aiding connection is
Equivalent inductance of parallel opposing connection is
Calculation:
Given,
L_{1} = 100 mH
L_{2} = 200 mH
M = 50 mH
Inductance of parallel aiding connection is
The inductance of a single layer coil of 50 turns is 5 mH. If the no. of turns is doubled, the inductance of the coil will become ______
Concept:
The formula for inductance of a coil is
L = μN^{2}A / l
N is no. of turns
A is the crosssectional area
l length of the solenoid
L ∝ N^{2} / l
L_{2}L_{1 }= (N_{2} / N_{1})^{2} x l_{1}l_{2}
Calculation:
Given:
L_{1} = 5 mH, N_{1} = 50 turns
Now, The number of turns is doubled i.e.
N_{2} = 2N_{1} = 100
l_{2 }= l_{1} = l
L_{2} = 20 mH
The _______ the permeability of the core material, the _______ will be the inductance.
Concept:
Inductance is the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance.
Therefore, the below equation will give inductance L as being proportional to the number of turns squared N^{2}
The inductance of a coil of wire is given by,
Where N is the number of turns
A is the crosssectional area
L is the length of the solenoid.
μ_{r} is the relative permeability of the core material
From the above relation, we observe that the inductance is directly proportional to the permeability of the material, i.e. higher the permeability of the core material, higher will be the inductance.
An ironcored coil has an inductance of 4 H. If the reluctance of the flux path is 100 AT/Wb, then the number of turns in the coil is:
Concept:
SelfInductance
Here, A = Area
N = Number of turns
l = Length of coil
magnetic flux
Reluctance s = N^{2} / L.......(i)
Calculation:
L = 4 H, s = 100 AT / Wb
From equation (i)
N = √4×100
= 20
A 500 W discharge lamp takes a current of 4 A at unity p.f. Find the inductance of a choke required to enable the lamp to work on 250 V, 50 Hz main.
Concept :
Choke coil (or ballast) is a device having high inductance and negligible resistance. It is used to control current in ac circuits and is used in fluorescent tubes. The power loss in a circuit containing choke coil is least.
(1) It consist of a Cu coil wound over a soft iron laminated core
(2) Thick Cu wire is used to reduce the resistance (R) of the circuit
(3) Soft iron is used to improve inductance (L) of the circuit
(4) The inductive reactance or effective opposition of the choke cost is given by X_{L} = 2πfL
(5) For an ideal choke coil r= 0, no electric energy is wasted i.e. average power P=0
6) In actual practice choke coil is equivalent to a RL circuit.
(7) Choke coil for different frequencies are made by using different substances in their core.
For low frequency L should be large thus iron core choke coil is used. For high frequency ac circuit, L should be small, so air cored choke coil is used.
Calculations:
P = I^{2 }R
X_{L} = 2 π f L
L= X_{L} / 2πf
Given
P = 500 W
I = 4 A
V = 250 V
f = 50 Hz
pf = unity
R = P / I^{2 }= 500 / 16 = 31.25 Ω
Z = V / I = 250 / 4 = 62.5 Ω
X_{L} = 54.125 Ω
L = X_{L }/ (2π f) = 0.1732 H
Self induction:
If L_{1} and L_{2} are the inductances of coil1 and coil2, then the total inductance of the magnetic circuit will be represented as
Concept:
Mutually Aiding:
Leq = L_{1} + L_{2} + 2M
L : selfinductance
M: Mutual inductance
Mutually Opposing:
Leq = L_{1} + L_{2} – 2M
An iron cored coil with 1000 turns generates a magnetic flux of 500 μwb in the core while carrying an electric current of 50 A. What will be the selfinductance of the coil?
Concept:
Selfinductance
Here, A = Area
N = Number of turns
l = Length of coil
magnetic flux
Calculation:
N= 1000, ϕ = 500 μwb, I = 50 A
R_{l = }reluctance
ϕ = (1000 × 50 ) / Rl
R_{l }= 10^{8}
L = 0.01 H
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