An inductor is a passive device that stores energy in a magnetic field when electric current flows through it. When an alternating current (AC) is applied to an inductor, the inductor can oppose or block the passage of the AC through it. This is due to the property of inductance, which is the measure of an inductor’s ability to store electrical energy in its magnetic field.
The voltage across an inductor can be measured as the amount of electromotive force (EMF) generated for the change of current. For example, if an inductor produces an EMF of 1 volt when the current changes at the rate of 1 ampere per second, we say the inductor has an inductance of 1 Henry.
AC Voltage and Inductor
We’re diving into how electric circuits work when we connect an AC voltage to an inductor. To understand this, let’s look at a circuit. In the circuit, we have an inductor and an AC voltage source, which we show with this symbol ~. This AC voltage source makes the voltage across the inductor change in a wavy pattern, like a sine wave. We can express this changing voltage with an equation.
The AC voltage, v, can be given as:
\(\displaystyle v = v_m \sin(\omega t)\)
- vm is the amplitude of the oscillating potential,
- ω is the angular frequency.
Using Kirchhoff’s loop rule, which states that the sum of the potential differences (voltages) around any closed loop or mesh in a network is always equal to zero, we can write:
\(\displaystyle v – L \frac{di}{dt} = 0\)
Here, L is the self-inductance of the inductor, and \(\displaystyle\frac{di}{dt}\) is the rate of change of current. From this, we get the equation for the change in current as a function of time:
\(\displaystyle\frac{di}{dt} = \frac{v}{L} = \frac{v_m}{L} \sin(\omega t)\)
This equation tells us how the current changes with time, but it doesn’t give us the current itself. Its amplitude is given by (vm/L). To find the current, we need to integrate this equation with respect to time.
\(\displaystyle \int{{di=\int{{\frac{{{{v}_{m}}}}{L}\sin \left( {\omega t} \right)}}}}\)
\(\displaystyle i=-\frac{{{{v}_{m}}}}{{\omega L}}\cos \left( {\omega t} \right)\)
This is because the integral of \(\displaystyle\sin(\omega t)\) with respect to (t) is \(\displaystyle -\frac{1}{\omega} \cos(\omega t)\). The integral of the sine function is negative cosine, so we get:
\(\displaystyle i = -\frac{v_m}{\omega L} \cos(\omega t) + C\)
So, this is the equation for the current through an inductor when an AC voltage is applied.
\(\displaystyle\boldsymbol { i = -\frac{v_m}{\omega L} \cos(\omega t)}\)
Here, C is the integration constant, which represents the initial current through the inductor at t = 0. The “integration constant” is a part of a mathematical equation that stays the same no matter what. In this case, it’s related to the current in the circuit. Since the source of the electricity we’re using swings back and forth equally around zero (like an AC voltage), the current does the same thing. That means there’s no steady, unchanging part of the current. So, in this situation, the integration constant ends up being zero.
Also Read: AC Voltage Applied To A Resistor
\(\displaystyle -\cos \left( {\omega t} \right)=\sin \left( {\omega t-\frac{\pi }{2}} \right)\)
\(\displaystyle i={{i}_{m}}\sin \left( {\omega t-\frac{\pi }{2}} \right)\)
Where \(\displaystyle i_m = \frac{v_m}{\omega L} \) is the amplitude of the current. The quantity ωL is analogous to the resistance and is called inductive reactance, denoted by XL: XL = ω L The amplitude of the current is, then
\(\displaystyle\boldsymbol{ i_m = \frac{v_m}{X_L} }\)
Inductive reactance is measured in ohms, just like resistance. It’s a bit like resistance in a circuit with only inductors. Just as resistance slows down the flow of electricity in a regular circuit, inductive reactance slows down the flow of current in a circuit with just inductors.
The amount of inductive reactance depends on how big the inductor is and how fast the current is changing. If the inductor is bigger or if the current is changing more quickly, the inductive reactance will be greater too.
The current is a sinusoidal function, just like the voltage, but it is shifted by 90 degrees (or \(\displaystyle\pi/2) radians\), which is a characteristic of inductors in AC circuits. This phase shift is due to the inductor’s resistance to changes in current.
A comparison of Eqs. \(\displaystyle v = v_m \sin(\omega t)\) and \(\displaystyle i={{i}_{m}}\sin \left( {\omega t-\frac{\pi }{2}} \right)\) for the source voltage and the current in an inductor shows that the current lags the voltage by π/2 or one-quarter (1/4) cycle. The figure shows the voltage and the current phasors in the present case at instant t1. The current phasor I is π/2 behind the voltage phasor V.
We notice that the current peaks occur after the voltage peaks by one-fourth of the time it takes for a full cycle to complete. This time delay is equivalent to one-fourth of the period of the wave \(\displaystyle \left[ {\frac{T}{4}=\frac{{{\pi }/{2}\;}}{\omega }} \right]\).
instantaneous power
Now, let’s see if this delayed current also consumes power, just like it does in a resistor. Let’s investigate further. The instantaneous power p supplied to any electrical device is given by the product of the instantaneous voltage v and the instantaneous current i. So, we have:
\(\displaystyle p = vi\)
Now, we know that in an inductor, the voltage and current are given by:
\(\displaystyle v = v_m \sin(\omega t)\)
\(\displaystyle i = i_m \sin(\omega t – \frac{\pi}{2})\)
Substituting these into the power equation gives us:
\(\displaystyle p = v_m \sin(\omega t) \cdot i_m \sin(\omega t – \frac{\pi}{2})\)
We can simplify this using the trigonometric identity for the product of sines:
\(\displaystyle\sin(A) \sin(B) = \frac{1}{2}[\cos(A – B) – \cos(A + B)]\)
So, we get:
\(\displaystyle p = \frac{1}{2} v_m i_m [\cos(\frac{\pi}{2}) – \cos(2\omega t – \frac{\pi}{2})]\)
Since \(\displaystyle \cos(\pi/2) = 0\), this simplifies to:
\(\displaystyle p = -\frac{1}{2} v_m i_m \cos(2\omega t – \frac{\pi}{2})\)
Now, we know from trigonometry that \(\displaystyle\cos(\theta – \frac{\pi}{2})\) is equivalent to \(\displaystyle\sin(\theta)\). This is because the cosine function leads the sine function by a phase of \(\displaystyle\frac{\pi}{2}\) radians (or 90 degrees).
So, we can rewrite the equation as:
\(\displaystyle \boldsymbol{ p_L = -\frac{i_m v_m}{2} \sin(2\omega t)}\)
The negative sign indicates that the power is being returned to the circuit half the time, which is a characteristic of reactive components like inductors and capacitors.
So, the average power over a complete cycle is
\(\displaystyle p_{L} = \left\langle -\frac{i_m v_m}{2} \sin(2\omega t) \right\rangle \)
\(\displaystyle p_{L} = -\frac{i_m v_m}{2} \left\langle \sin(2\omega t) \right\rangle = 0 \)
Since the average of sin (2ωt) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero.
Examples
Example 1: An AC voltage source with a frequency of 50 Hz is applied to an inductor with an inductance of 0.02 H Calculate the reactance of the inductor.
Solution:
The reactance XL of an inductor in an AC circuit is given by the formula:
\(\displaystyle X_L = 2\pi f L \)
where f is the frequency and ( L ) is the inductance. Given that f = 50 Hz and L = 0.02 H we can calculate the reactance:
\(\displaystyle X_L = 2\pi \times 50 \times 0.02 \)
\(\displaystyle X_L = 2\pi \, \Omega \)
Therefore, the reactance of the inductor is \(\displaystyle 2\pi \, \Omega \).
Example 2: An AC voltage source with a frequency of 100 Hz and an effective voltage of 10 V is applied to an inductor. If the current lags the voltage by (π/4) radians, calculate the inductance of the inductor.
Solution:
In an AC circuit, the relationship between voltage V, current I, frequency f, and inductance L, for an inductor is given by:
\(\displaystyle V = IXL \)
Given that the current lags the voltage by (π/4) radians, we know that the phase angle (θ = -π/4). Therefore, we have:
\(\displaystyle X_L = \frac{V}{I} = \frac{V}{V/X_L} = X_L^2 = \frac{V^2}{V} = V \)
\(\displaystyle X_L = 10 \, \Omega \)
We use the formula for reactance:
\(\displaystyle X_L = 2πfL \)
Substituting the given values, we can solve for L:
\(\displaystyle 10 = 2π \times 100 \times L \)
\(\displaystyle L = \frac{10}{200π} \)
\(\displaystyle L = \frac{1}{20π} \, \text{H} \)
Therefore, the inductance of the inductor is \(\displaystyle\frac{1}{20π} \, \text{H} \).
Example 3: An AC voltage source with an effective voltage of 20 V and a frequency of 60 Hz is applied to an inductor. If the current through the inductor is 1 A, determine the inductance of the inductor.
Solution:
Given parameters:
- Effective Voltage Veff = 20 V
- Frequency f = 60 Hz
- Current I = 1 A
We know that the reactance XL of an inductor is given by:
\(\displaystyle X_L = \frac{V_{\text{eff}}}{I} \)
And the formula for reactance of an inductor is:
\(\displaystyle X_L = 2πfL \)
Equating both expressions for XL we get:
\(\displaystyle \frac{V_{\text{eff}}}{I} = 2πfL \)
\(\displaystyle L = \frac{V_{\text{eff}}}{2πfI} \)
Substituting the given values:
\(\displaystyle L = \frac{20}{2π \times 60 \times 1} \)
\(\displaystyle L = \frac{20}{120π} \)
\(\displaystyle L = \frac{1}{6π} \, \text{H} \)
Therefore, the inductance of the inductor is \(\displaystyle \frac{1}{6π} \, \text{H} \).
Example 4: An AC voltage source with a frequency of 50 Hz is applied to an inductor. If the reactance of the inductor is 10 Ω, calculate the inductance of the inductor.
Solution:
Given parameters:
- Frequency f = 50 Hz
- Reactance XL = 10 Ω
We know that the reactance XL of an inductor is given by:
\(\displaystyle X_L = 2πfL \)
Solving for inductance L:
\(\displaystyle L = \frac{X_L}{2πf} = \frac{10}{2π \times 50} = \frac{10}{100π} \)
\(\displaystyle L = \frac{1}{10π} \, \text{H}\)
Therefore, the inductance of the inductor is \(\displaystyle\frac{1}{10π} \, \text{H} \).
Example 5: An AC voltage source with an effective voltage of 30 V and a frequency of 100 Hz is applied to an inductor. If the inductance of the inductor is 0.02 H, calculate the current through the inductor.
Solution:
Given parameters:
- Effective Voltage Veff = 30 V
- Frequency f = 100 Hz
- Inductance L = 0.02 H
We know that the reactance XL of an inductor is given by:
\(\displaystyle X_L = 2πfL \)
And the formula for current (I) through the inductor is:
\(\displaystyle I = \frac{V_{\text{eff}}}{X_L} \)
Substituting the given values:
\(\displaystyle X_L = 2π \times 100 \times 0.02 = 4π \, \text{Ω} \)
\(\displaystyle I = \frac{30}{4π} \)
Therefore, the current through the inductor is
\(\displaystyle\frac{30}{4π} \) amperes.