23.5 ALTERNATING CURRENTS THROUGH INDUCTORS
23.5.1 Phase Difference between I and V for Pure Inductors
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| Figure 23.21 The figure shows a pure inductor connected to a sinusoidal voltage source, producing a sinusoidal current in the circuit. |
A pure inductor with inductance L is connected to a sinusoidal voltage supply of voltage V, as shown in Figure 23.21. The voltage produces a sinusoidal current I which is given by
I = I0 sin ωt
where I0 is the maximum current. Since the current is changing with time, self induction occurs in the inductor, producing a back emf Eb. The back emf is given by
Eb = −L dI/dt
= −L d/dt (I0 sin ωt)
= −ωLI0 cos ωt
Since the inductor is assumed to be ideal, we have
applied voltage = – back emf
VL = –(–ωLI0 cos ωt)
= ωLI0 cos ωt
= (ωLI0) sin (ωt + ½π)
Let us make a comparison:
I = I0 sin ωt
VL = (ωLI0) sin (ωt + ½π)
Notice that I has a ‘sin ωt’ term whereas VL has a ‘sin (ωt + ½π)’ term. This means that I is changing not in step with VL. In fact, I is out of step with VL by ½π radian. We express this fact by saying that there is a phase difference of ½π radian between VL and I, with VL leading I.
23.5.2 I–t graph and VL–t graph for Pure Inductor
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| Figure 23.22 The figure shows the I–t and Vₗ–t graphs along with a phasor diagram, illustrating the phase relationship between current and voltage in an inductor. |
We can draw one I–t graph and one VL–t graph using the I and VL functions given above. Figure 23.22 shows these two graphs. The phasor diagram is incorporated in the graphs.
We notice that:
- (a) there are two arrows, one representing VL and the other representing I.
- (b) the two arrows are 90° apart, with VL rotating ahead of I. This indicates that VL leads I by ½π radian.
- (c) I = 0 when VL is maximum, and I is maximum when VL = 0. In other words, I varies out of step with VL. This is the main characteristic of two quantities varying out of phase by 90°.
- (d) It is possible to get values where (+VL, +I), (+VL, –I), (–VL, +I), (–VL, –I). This means that the product VLI for a pure inductor could have a positive or a negative value.
23.5.3 Inductive Reactance
1 Meaning
When alternating current flows through a pure inductor, the current experiences some opposition although there is no resistance in the circuit. This opposition, which arises due to the inductance possessed by the inductor, is the inductive reactance.
2 Magnitude
We have
VL = (ωLI0) cos ωt
When the voltage becomes maximum, we have
cos ωt = ±1
and so,
V0 = ωLI0
or,
V0 / I0 = ωL
The ratio V0/I0 is the inductive reactance, XL. Hence, we have
XL = V0 / I0 (inductive reactance)
V0 = I0XL
XL = ωL
where
ω = 2πf
and f is the frequency of the alternating current or voltage.
We have
V0 = √2 Vrms
I0 = √2 Irms
Hence,
XL = Vrms / Irms
3 Unit
The unit of inductive reactance is the ohm.
4 Graph of XL – f
XL = 2πfL
XL ∝ f
XL ∝ L
The graph in Figure 23.23 shows how XL varies linearly with f.
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| This graph shows that the inductive reactance, $X_L$, increases linearly with frequency, f. Since $X_L= 2\pi fL$, the relationship is directly proportional, meaning that as the frequency increases, the opposition offered by the inductor to the alternating current also increases. |
EXAMPLE 23.17
What must be done in order to double the inductive reactance of a pure inductor?
Answer
XL ∝ f
XL ∝ L
Hence, we can:
- (a) double the frequency of the current flowing through the inductor
- (b) double the inductance of the inductor
EXAMPLE 23.18
A current I which flows through an ideal inductor with inductance of 200 mH is given by
I = (2.0) sin (100πt)
(a) Determine
(i) the inductive reactance of the inductor
(ii) the maximum voltage that can appear across the inductor
(b) Write down an expression for the voltage V which is applied across the inductor:
Answer
(a) (i) Given
ω = 100π
XL = ωL
= (100π)(0.200) = 62.8 Ω
(ii)
V0 = I0XL
= (2.0)(62.8) = 126 V
(b)
VL = V0 sin (ωt + ½π)
= (126 V) sin (100πt + ½π)
EXAMPLE 23.19
A coil with inductance 100 mH and negligible resistance is connected to a sinusoidal voltage of frequency 50 Hz. If the current flowing through the coil is 8.0 A (rms), determine:
- (a) the inductive reactance
- (b) the rms voltage applied across the coil
Answer
(a)
XL = 2πfL
= 2π(50)(0.100) = 31.4 Ω
(b)
Vrms = Irms XL
= (8.0)(31.4) = 251 V
23.5.4 Power in Inductive Circuit
1 Instantaneous power
Suppose that a current I given by
I = I0 sin ωt
passes through a pure inductor. Then the voltage VL which is applied across the inductor to produce this current is given by
VL = V0 cos ωt
since V leads I by ½π radian. The instantaneous power P delivered to the inductor is given by
P = IVL
= (I0 sin ωt)(V0 cos ωt)
= I0V0 (sin ωt)(cos ωt)
= ½ I0V0 sin (2ωt)
The maximum power available at an instant is ½ I0V0.
%20Graph%20in%20an%20Inductive%20Circuit.webp) |
| Figure 23.24 The figure shows how instantaneous power varies with time, alternating between positive and negative values over one cycle. |
Figure 23.24 shows a graph of P against time t. Notice that when the current has gone through one cycle,
- (a) the power will have gone through two cycles
- (b)
- (i) the power has two regions of positive values and two regions of negative values
- (ii) area of positive region = area of negative region
- (iii) total area for one current cycle = 2 positive areas + 2 negative areas = 0. This implies that the average power is zero.
2 Average power
The average value of sin (2ωt) is zero. This means that the average power Pav is
Refer to the power curve shown in Figure 23.24 above. We notice that the positive power is cancelled by the negative power, producing a net power of zero over every one cycle of the power curve.
Reasons:
- (a) During one half cycle of positive power, the inductor absorbs electrical energy. This energy is converted to magnetic energy stored in the magnetic field produced by the current flowing through the inductor.
- (b) The positive power is followed by one half cycle of negative power. Now the magnetic field is shrinking and collapsing. The magnetic energy is diminishing with time. During this half cycle, the magnetic energy is converted into electrical energy and returned back to the circuit.
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(c) Over one power cycle, the energy of the system undergoes transformation from one form of energy into another, without any loss to the surroundings:
electrical energy ⇌ magnetic energy - (d) Since the ideal inductor has no resistance, no energy is dissipated in the form of heat.
(Note: We are assuming that there are no electromagnetic waves generated as the electrons oscillate in the circuit.)
⚡ Power Transmission Systems
Inductors are widely used in power transmission lines to control current and reduce sudden changes. They help stabilize the flow of alternating current and protect equipment from voltage spikes.
🔊 Audio Equipment (Speakers & Filters)
Inductors are used in crossover circuits in speakers to filter frequencies. They allow low-frequency signals to pass while blocking high-frequency signals.
📡 Radio and Communication Circuits
Inductors are essential in tuning circuits (LC circuits) used in radios, TVs, and wireless communication. They help select specific frequencies from signals.
🔌 Transformers
Transformers rely on inductance to transfer energy between coils. They are used to step up or step down voltage in electrical systems.
⚙️ Energy Storage (Magnetic Field)
Inductors store energy temporarily in their magnetic field when current flows through them. This energy is released back into the circuit when current decreases.
💡 Power Factor Correction
Inductors affect the phase difference between voltage and current. They are used in power systems to manage power factor and improve efficiency.
The concept of inductive reactance (XL) explains how inductors oppose alternating current, with its value depending on both frequency and inductance. As frequency increases, the opposition to current also increases.
Additionally, although instantaneous power varies with time, the average power over a complete cycle is zero. This indicates that energy is not dissipated but temporarily stored and returned to the circuit. Key Insight: Inductors do not consume energy — they store and release it, making them essential components in AC circuits, power systems, and electronic devices.
Inductive reactance is the opposition offered by an inductor to alternating current. It is given by XL = 2πfL.
Voltage leads current by 90° because the inductor produces a back emf that opposes the change in current, creating a phase difference.
Energy is stored in the magnetic field during one half cycle and returned during the next, resulting in zero net energy consumption.
It depends on frequency (f) and inductance (L).
Inductors are used in transformers, filters, power supplies, radios, and audio systems.
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