Power in AC Circuits: Do you know DC and AC are two different entities? Do you know that direct current (dc) sources and circuits with dc sources do not change direction with time? But the magnitude of voltages and currents vary with time. The electric mains supply a voltage that changes like a sine function with time in our homes and offices. This changing voltage is known as an alternating voltage(ac voltage), and the current in this circuit is called the alternating current(AC).

As ac voltages can be easily and efficiently converted from one voltage to another using a transformer and transmitted economically over long distances. So, an ac voltage is preferred over dc voltage. In this article, we will learn about power consumption in ac circuits and its derivation. We will also discuss power dissipation for different circuits such as resistive circuits, purely inductive or capacitive circuits, and LCR series circuits.

LRC Circuit

Suppose a voltage \(V\) is applied to an \(LCR\) circuit, where is given by:

\(V=V_{m} \sin \omega t\)

Then the current will be,

\(i=i_{m} \sin (\omega t – \varphi)\)

Where, \(V_{m}=\) Voltage Amplitude, 
\(i_{m}=\) Current Amplitude, 
\(\omega=\) Angular frequency, 
\(\varphi=\) Phase Constant

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Now, the relation between current Amplitude and voltage amplitude is given as: 

\(i_{m}=\frac{V_{\mathrm{m}}}{Z}\)

Where, \(Z=\) Impedance of circuit 

Impedance and Impedance Diagram

The circuit shown in the above figure shows that the resistor, inductor, and capacitor are in series. Impedance is the combination of resistance and reactance. It can be represented by \(Z\), and its unit is the ohm. From the figure, the Impedance diagram is a right-triangle that represents a combination of resistance \((R)\) and reactance \((X)\) with Impedance \((Z)\) as its hypotenuse.

Where reactance \((X)\) is a combination of Inductive \(\left(X_{L}\right)\) and capacitive \(\left(X_{C}\right)\)

\(Z=\sqrt{\left(X_{c}-X_{L}\right)^{2}+R^{2}}\)

\(\tan \varphi=\frac{X_{c}-X_{L}}{R}\)

\(\varphi=\left(\frac{X_{C}-X_{L}}{R}\right)\)

Where,

\(X_{C}=\frac{1}{\omega C}, X_{C}=\) Capacitive Reactance  

\(X_{L}=\omega L, X_{L}=\)Inductive Reactance

Power in AC Circuit

In the case of a steady current, the rate of doing work is given by

\(P=V i\)

In an alternating circuit,  both current and voltage vary with time, and they differ in time. So, we cannot use \(P =Vi\) directly. Suppose in an AC; the voltage is leading the current by an angle \(\varphi\). Then, we can write.

\(V=V_{0} \sin \omega t\) And

\(i=i_{0} \sin (\omega t-\varphi)\)

Then, we can write the instantaneous value of power as:

\(P=V i=V_{0} i_{0} \sin \omega t \sin (\omega t-\varphi)\)

\(P=V_{0} i_{0}\left[\sin ^{2} \omega t \cos \varphi-\frac{1}{2} \sin 2 \omega t \sin \varphi\right]…\left( 1 \right)\)

Now, the average rate of doing work (power) in one cycle will be:

\({\langle P\rangle _{{\text{one}}\;{\text{cycle}}}} = \frac{{\mathop \smallint \nolimits_0^{T = \frac{{2\pi }}{\omega }} Pdt}}{{\mathop \smallint \nolimits_0^{T = \frac{{2\pi }}{\omega }} dt}}\) …….(2)

Substituting the value of \(P\) from Eq. \((1)\) in Eq. \((2)\) and then integrating it with proper limits, we get:

\(\langle P\rangle_{\text {one cycle }}=\frac{1}{2} V_{0} i_{0} \cos \varphi=\frac{V_{a}}{\sqrt{2}} \cdot \frac{i_{0}}{\sqrt{2}} \cos \varphi\)

Or,

\({\langle P\rangle_{{\text{one}}\,{\text{cycle}}}} = {V_{rms}}{i_{rms}}\cos \;\varphi \)

It can also be written as,   

\(P = {I^2}Z \cos \varphi \,\,\,…\left( 4 \right)\)

So, the average power dissipated depends on the voltage and current. It also depends on the cosine of the phase angle \(\varphi\) between them. Where \( \cos \varphi\) is called the power factor. Let us discuss the following cases:

Case (i) Resistive Circuit: It is called resistive if the circuit contains only pure \(R\). In that case, \(\varphi=0, \cos \varphi=1\). There is maximum power dissipation.

Case (ii) Purely Inductive or Capacitive Circuit: Hence, the phase difference between voltage and current is \(\frac{\pi}{2}\). Suppose the circuit contains only an inductor or capacitor. Therefore no power is dissipated even though a current is flowing in the circuit as \(\cos \cos \varphi=0\) We call this current a wattless current.

Case (iii) LCR Series Circuit: Here,  the power dissipated is given by Eq. \((3)\) and Eq \((4)\), where \(\varphi  = {\tan ^{ – 1}}\left( {\frac{{{X_C} – {X_L}}}{R}} \right)\).
So, \(\varphi\) may be non-zero in an \(RL\) or \(RC\) or \(RLC\) circuit.

Case (iv)At resonance in \(LCR\) circuit: At resonance: \(X_{C}-X_{L}=0\) and \(\varphi=0 .\). Therefore, \(\cos \varphi=1\) and hence the power dissipated can be written as \(P=I^{2} Z=I^{2} R\). Hence, maximum power is dissipated in a circuit at resonance through \(R.\)

Power Factor

The power factor is defined as the ratio of the real power flowing to the load to the apparent power in the ac circuit. It is a dimensionless quantity and in the closed interval of  \(-1\) to \(1\).

Power Factor \(=\frac{\text { True Power }}{\text { Apparent Power }}\)

Since, 

\(P=V_{r m s} i_{r m s} \cos \varphi\)

Here, the term \(\cos \varphi\) is known as the power factor. For example, it is said to be leading if current leads voltage, lagging if current lags the voltage. Thus, a power factor of \(0.5\) lagging means the current lags the voltage by \(60^{\circ}\left(0.5=60^{\circ}\right)\). The product of \(V_{\mathrm{rms}}\) and \(i_{\text {rms }}\) gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor. Thus,

Apparent power \(=V_{r m s} i_{r m s}\)

And,

True power \(=\) apparent power \(\times\) power factor

Phasor Diagram for Power Factor

As we know, that \(=V I \cos \varphi\), where \(\cos \varphi\) is the power factor. If \(\cos \varphi\) is small, we must increase the current accordingly to supply a given power at a given voltage. But this will lead to large power loss \(\left(I^{2} R\right)\) in transmission.

Now, let us suppose in a circuit, current \((I)\) lags the voltage by an angle \(\varphi\) . Then

power factor,

\(\cos \varphi=\frac{R}{Z}\)

By making \(Z\) tend to \(R\), we can improve the power factor (tending to \(1)\). With the help of a phasor diagram, as shown in the figure below, let us understand how this can be achieved. Let us resolve \(I\) into two components one is along with the applied voltage that is \(I_{p}\), and the other perpendicular to the applied voltage that is \(I_{q}\)

Where \(I_{q}\) is known as a wattless component because there is no power loss here, and \(I_{p}\) is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

Power in AC Circuits– Solved Examples

Q.1. The following data are given when a sinusoidal voltage is applied to a series LCR circuit whose peak value is given as \(283\;{\rm{V}}\) and frequency is given as \(50\;{\rm{Hz}}\) in which the value of \(R = 3\,\Omega,\;L = 25.48\;{\rm{mH}}\) and \(C= 796\;{\rm{μF}}\). Then find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.
Ans:
(a) Impedance
To find the impedance of the circuit, we first calculate \(X_{L}\) and \(X_{c}\)
\(X_{L}=2 \pi v L\)
\(\Rightarrow X_{L}=2 \times 3.14 \times 50 \times 25.48 \times 10^{-3}\,\Omega\)
\(X_{c}=\frac{1}{2 \pi v C}\)
\(\Rightarrow X_{c}=\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-1}}=4\,\Omega\)
Therefore, the impedance will be:
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{3^{2}+(8-4)^{2}}\)
\(\Rightarrow Z=5\,\Omega\)
(b) Phase difference,
\(\varphi  = {\tan ^{ – 1}}\left( {\frac{{{X_C} – {X_L}}}{R}} \right)\)
\(\Rightarrow \varphi= {\tan ^{ – 1}} \left(\frac{4-8}{3}\right)=-53.1^{\circ}\)
(c) The power dissipated in the circuit is given by :
\(P=I^{2} R\)
Now,
\(I=\frac{I_{m}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{283}{5}\right)=40\,{\rm{A}}\)
Therefore,  
\(P=(40 A)^{2} \times 3 \Omega=4800\,{\rm{W}}\)
(d) Power factor
Power factor \(=\varphi\)
\(\varphi=\cos 53.1^{\circ}=0.6\)

Summary

We can get the average ac power by multiplying the rms values of current and voltage. In an ac circuit, we can get the phase angle between the source voltage and the current by dividing the resistance to the impedance. The range of power factor lies from \(–1\) to \(1\).

For a series, \(RLC\) circuit driven by voltage, \(V=V_{m} \sin \omega t\) the current is given by \(I=I_{m} \times \sin (\omega t+\varphi)\) where, \(I_{m}=V_{m} / Z\) and \(Z=\sqrt{\left(X_{c}-X_{L}\right)^{2}+R^{2}}\) Where \(Z\) is called the impedance of the circuit, \(X_{L}=\omega L\) is called inductive reactance, and \(X_{C}=1 / \omega C\) is called capacitive reactance.

Over a complete cycle, the average power loss is given by \(P=V I \cos \varphi\) Where the term cos is called the power factor. In a purely inductive or capacitive circuit, no power is dissipated even though a current is flowing in the circuit as, \(\cos \varphi=0\) In such cases, the current is referred to as a wattless current.

Frequently Asked Questions (FAQs)

Q.1. Define the term ‘wattless current.
Ans: The component of the circuit current due to which the power consumed in the circuit is zero, then this component is known as wattles current.

Q.2. What do you mean by the term rms value of the current? Find its relation with the peak value?
Ans: RMS value of the current is that value of Alternating Current (AC) over a complete cycle that would generate the same amount of heat in a given resistor caused by the steady current in the same resistor and at the same time during a complete cycle. Let the peak value of the current be \(I_{0}\) then, 
\(I_{r m s}=\frac{I_{2}}{\sqrt{2}}\)

Q.3. Distinguish between resistance, reactance, and impedance.
Ans: The opposition offered by a resistor \(R\) when a current flow in an ac circuit is known as the resistance; The opposition offered by an inductor \(L\) or by a capacitor \(L\) or both is termed as the reactance, while the opposition offered \(L-R, C-R,\) or \(L-C-R\) circuits is termed as the impedance. 

Q.4. How do you find the power in an AC circuit?
Ans: The instantaneous electric power in an \(AC\) circuit is given by \(P=VI\), where \(v\) and \(I\) are the instantaneous voltage and current, and averaging this power over a complete cycle gives the average power.

Q.5. What is instantaneous and average power?
Ans: The instantaneous power is the power at a given instant. Average power is that power that is averaged over a cycle or number of cycles. 

We hope this detailed article on Power in AC Circuits helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!