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AC generator is a device that converts mechanical energy into electrical energy. The AC generators works on the principle of Faraday’s Law of Electromagnetic Induction. In an AC generator, the mechanical energy is supplied through turbines and combustion engines. The generator then converts this mechanical energy into electrical power in the form of alternating voltage and current. AC generator is also known as AC Dynamo or Alternator.
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Key Terms: AC Generator, DC Generator, Electromagnetic Induction, Magnetic Field, Electromotive Force, Voltage, Current, Combustion engines, Electrical power, Alternator
What is an AC Generator?
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AC generators, also referred to as alternators, are machines that converts available mechanical energy into electrical energy. This can be achieved in two ways:
- By rotating a conducting coil during a static magnetic flux
- By rotating the magnetic flux that comprises the stationary conductor.
The preferred method is to use a rotating magnet inside a stationary coil. This is because it is easier to extract the induced alternating current from a stationary armature coil than from a rotating coil. The rotating magnet inside the coil generates a magnetic field which induces a current in the wire. The generated EMF relies upon the amount of armature coil turns, the strength of the magnetic field, and also on the speed of the rotating field.
AC Generator
Types of AC GeneratorsThere are two main types of AC generator:
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Parts of an AC Generator
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An AC generator is made up of the following parts and each part perform its specific function:
- Field – The field comprises coils of conductors that gets voltage from the source and produce magnetic flux. The magnetic flux within the field cuts the armature to produce a voltage. This voltage is considered as the output voltage of the AC generator.
- Armature – Armature is the part of the AC generator where the voltage is produced. It consists of coils of wire that are capable of carrying the full-load current of the generator.
- Prime Mover – Prime mover is the component which is used to drive the AC generator. The prime mover could be a diesel engine, a turbine, or a motor.
- Rotor – It is the rotating component of the generator. The rotor is operated by the prime mover of the generator. Depending on the type of generator, this component may be considered as the armature or the field. When the voltage output is generated there, the rotor will be the armature and it will be the field if the field excitation is applied there.
- Stator – The stationary part of the AC generator is stator. Like the rotor, depending on the kind of generator, this component could also be the armature or the field. The stator will be the armature if the voltage output is generated there and field if the field excitation is applied there.
- Slip Rings – Slip rings form the electrical connections that transfer power to and from the rotor of an AC generator. They are structured to manage the flow of electric current from a stationary device to a rotating one.
Parts of an AC Generator
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| Magnetic Properties of Material | Lenz’s law | Solenoid Engine |
| Types of generators | Electromagnetic Damping | Magnetic Induction Formula |
Working of an AC Generator
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AC generator works on the principle of Faraday’s Law of Electromagnetic Induction. Whenever a conductor is placed within the lines of the magnetic flux, an emf is induced. This emf causes the current to pass through the conductors.
- When the armature rotates between the poles of the magnet upon an axis which is perpendicular to the magnetic flux, the flux linkage of the armature changes.
- This is because an emf is induced within the armature. This generates an electrical current that flows through the galvanometer, slip rings and brushes.
- The galvanometer fluctuates between positive and negative values which indicates that there is an alternating current flowing through the galvanometer.
- The direction of the induced current can be determined using Fleming’s Right Hand Rule.
Faraday Laws of Electromagnetic InductionThe two laws of electromagnetic induction proposed by Faraday are as follows:
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Advantages of AC Generators
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The advantages of an AC generator over the DC generator are as follows:
- AC generators can be easily stepped up and stepped down through transformers.
- Their transmission link size is thinner due to the step-up feature.
- The losses in an AC generator are less than that in the DC generator.
- The size of the AC generators is comparatively smaller than DC generators.
Things to Remember
- An AC generator is a device that converts mechanical energy into electrical energy.
- The principle behind the working of the AC generators is the Faraday’s Law of Electromagnetic Induction.
- When the rotating component of the AC generator rotates in the magnetic field, an emf is induced. This induced emf generates current.
- AC generators are preferred over DC generators because they can be stepped up and down easily with the help of transformers.
- As compared to the DC generator, the losses in the AC generator are low.
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Sample Questions
Ques. State which one of the following quantities is zero for an ideal inductor connected across a sinusoidal ac voltage source. (1 Mark) [Comptt. All India 2016]
(i) instantaneous
(ii) average power over full cycle of the ac voltage source
Ans. The average power over full cycle of the ac voltage source is zero when connected with an ideal inductor.
Ques. Plot a graph that shows variation of capacitive reactance with the change in the frequency of AC source. (1 Mark) [Comptt. All India 2015]
Ans. The graph showing a variation of XC capacitive reactance with the change in frequency of AC source is:
Ques. In a circuit of coil of resistance 2Ω , the magnetic flux changes from 2 Wb to 10 Wb in 0.2 sec. What is the charge flow in the coil during this time? (2 Marks)
Ans. The relation between the rate of change of current and the flux is given by,
\(\frac{dQ}{dt} = \frac{-1 d\phi}{R dt}\)
\(\frac{dQ}{dt} = \frac{-(10 - 2)}{2}\) = 4 C
Ques. A closely wound flat circular coil of 25 turns of wire has diameter of 5 cm which carries current of 4 A. What will be the flux density at the centre of the coil? (2 Marks)
Ans. Flux Density, B = \(\frac{\mu_0 i}{2r} \times n\)
= 4\(\pi\) × 10-7 × 4 × 25 / (2 × 5/100)
= 1.256 × 10-3 Tesla
Ques. An inductor L of inductance XL is connected in series with a bulb B and an ac source. How can the brightness of the bulb change when (i) number of turns in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in the circuit in series. Justify your answer in each case. (3 Marks) [CBSE 2015]
Ans. (i) Given, net resistance in the circuit,
\(Z = \sqrt {{X_L}^2 +R^2}\)Inductance is,
L decreases as number of turns decreases.
Inductance is given by XL = ωL
Hence, XL will also decrease thereby, reducing the net resistance in the circuit. Therefore, current increases and brightness of the bulb increases as well.
(ii) When a soft iron rod is inserted in the circuit, L increases. Hence, the inductive reactance also increases. The net resistance increases and the flow of current decreases. Therefore, the brightness of the bulb also will decrease.
(iii) When a capacitor of reactance XL = XC is connected in series with the circuit net resistance becomes,
\(Z = \sqrt {(X_L-X_C)^2 +R^2}\)
Where R = resistance of the bulb
Here, we have Z = R which is a condition of resonance
Maximum current will flow through the circuit at resonance which is why the brightness of the bulb will increase.
Ques. (a) In a series LCR connected across an ac source of a variable frequency obtain the expression for its impedance and draw a plot that shows its variation with frequency of the ac source.
(b) State the phase difference between the voltages across the inductor and capacitor at resonance in the LCR circuit.
(c) Explain why: When an inductor is connected to 200 V dc voltage, a current of 1 A flows through it. When the same inductor is connected to 200 V (50 Hz ac source, only 0.5 A current flows. Also calculate the self-inductance of the inductor. (5 Marks) [CBSE 2019]
Ans. a) In the series LCR connected across an ac source of variable frequency,
Voltage of ac source is V = Vm sin ωt and current is I =Im sin ωt.
Voltage drop across resistor R be,
VR = ImR
The voltage is in the same phase of the current. And the voltage drop across inductor is VL = Im XL
Here XL = inductance
L X L = ω
Now, the voltage leads the current by π/2
The voltage drop across capacitor, Vc = Im XC
Here, XC = capacitive reactance: XC = 1/ ωc
The voltage lags behind the current by π/2
The resulting voltage will be vector sum of all voltage which is represented by OF
This is the required expression for impedance
At resources frequency, fr we have XL = XC and Z = R
b) At resonance, the phase angle of circuit = 0
Thus, the voltage across the inductor and capacitor, at resonance, are the same at any instant and cancel out each other. Therefore, the voltage drop across the LCR circuit is due to the voltage drop across resistance R. This is the reason why the phase difference between voltages across inductor and capacitor at resonance is 180°.
c) When the inductor is connected to 200 V dc voltage, the inductor acts as a resistor where there is no inductive reactance.
The resistance of coil can be get by using Ohm’s law, V =IR
Therefore, R =V/I
Given, A current flows when the inductor is connected to 200 V dc
\(R = \frac{200}{1}\)= 200 ΩWhen the same inductor is connected to 200 V, 50 Hz ac source, the inductive reactance causes change in total impedance of the coil because of which current changes.
Inductive reactance, XL - WL (frequency is 50 Hz)
The angular frequency is,
Hence, the self inductance of the inductor L is (√2/π)H
Ques. A device ‘X’ is connected to an ac source V =V0 sin ωt. The graph diagram given below shows the variation of voltage, current and power in one cycle. (5 Marks) [CBSE 2017]
(a) Identify the device ‘X’
(b) Justify with your answer that which of the curves A, B and C represent the voltage, current and power consumed in the circuit.
(c) How does its impedance vary with the frequency of the ac source. Show graphically.
(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage.
Ans. (a) The device X represents a pure capacitor.
(b) Curve A represents power, P = VI, where the amplitude is equal to the multiplication of V and I curves’ amplitudes. Curve B, which represents voltage is a sine curve and curve C is a cosine curve which represents the current.
The full cycle of the graph comprises the two positive and two negative symmetrical areas which is why the average power consumed in the circuit is zero.
(c) The ac impedance of a capacitor is called Reactance and the capacitor circuits are more commonly known as Capacitive Reactance. The graph shows variation of a capacitive reactance with frequency.
z = xc = 1/ ωc
xc = 1/ 2 π fc
vc ∝ 1/ fc
(d) Voltage in circuit, V = q/c = E = E0 sin ωt
The current is leading the ac voltage by an angle of π/2.
Ques. (i) An ac source of voltage V = V0 sin ωt is connected to a series of combination of L, C and R. using the phasor diagram, obtain an expression for impedance of a circuit and the phase angle between voltage and current. Find the condition when the current will be in phase with the voltage. What is the circuit in this condition known as?
(ii) In a series LR circuit, XL = R and the power factor of the circuit is P1. When a capacitor with capacitance C, such that XL = XC is put in series, the power factor becomes P2.
Calculate P1/ P2. (5 Marks) [CBSE 2016]
Ans. (i) Voltage of the source is V = V0 sin ωt
Let the current of source be, I = Io sin ωt
Maximum voltage across R, VR = Vo R, represented along OX
Maximum voltage across L, VL = Io XL, represented along OY which is 90° ahead of Io
Maximum voltage across C, VC = Io XC, represented along OC which is lagging behind Io by 90°
Hence, the reactive voltage is VL - VC represented by OB’
The vector sum of VR, VL and VC is the outcome of OA and OB’ and represented along OK.
OK = Vo =
Ques. In the given circuit, calculate
(a) if the power factor of the circuit is unity, then the capacitance of the capacitor
(b) the Q factor of the circuit. State the significance of the Q factor in ac circuit. Given the angular frequency of the ac source to be 100/s. Calculated the average power dissipated in the circuit. (5 Marks)
Ans. 
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