Power in AC Circuits

Alternating Current (AC) circuits are widely used in power transmission and household systems. Unlike DC circuits, where current is constant, AC circuits involve continuously varying current and voltage.

To study efficiency and performance of devices connected to AC supply, it is essential to understand power in AC circuits. This includes definitions, formulas, derivations, and practical applications.

1.0Power in AC Circuits Definition

Power in AC circuits refers to the rate at which electrical energy is transferred or converted into another form (heat, mechanical work, etc.) when alternating voltage and current are applied. Unlike DC circuits, the presence of phase difference between current and voltage plays a significant role in determining actual power consumed.

• In general:
• • Real Power (P): Useful, consumed power.
  • Reactive Power (Q): Power stored and released by reactive components.
  • Apparent Power (S): Total supplied power.

2.0Power in AC Circuits

The rate of doing work or the amount of energy transferred by a circuit per unit time is known as power in AC circuits. It is used to calculate the total power required to supply a load.

Instantaneous Power

$$\begin{gathered} As{P}_{\text{inst}}=VI=(V_0\sin \omega t),[I_0\sin (\omega t+\varphi )] \\ =V_0{I}_0\sin \omega t\sin (\omega t+\varphi )=\frac{V_0{I}_0}{2},[2\sin \omega t\sin (\omega t+\varphi )] \\ \text{Hence } \\ {P}_{\text{inst}}=\frac{V_0{I}_0}{2},[\cos \varphi -\cos (2\omega t+\varphi )] \\ \text{Since }2\sin A\sin B=\cos (A-B)-\cos (A+B) \end{gathered}$$

Note:Therefore frequency of power fluctuation is twice the frequency of applied a.c source.

$$\begin{gathered} \text{Average Power:} \\ {P}_{av}=\frac{{\int }_0^T(V_0\sin \omega t)(I_0\sin (\omega t+\varphi )),dt}{{\int }_0^{T}dt}=\frac{V_0{I}_0}{T}\left[\cos \varphi {\int }_0^T{\sin }^2\omega t,dt\frac{\sin \varphi }{2}{\int }_0^T{\sin }^2\omega t,dt\right] \\ {P}_{av}=V_0{I}_0\left[\cos \varphi \frac{{\int }_0^T{\sin }^2\omega t,dt}{T}\frac{\sin \varphi }{2}\frac{{\int }_0^T{\sin }^2\omega t,dt}{T}\right]=V_0{I}_0\left[\cos \varphi \times \frac{1}{2}+0\right] \\ {P}_{av}=\frac{1}{2}{V}_0{I}_0\cos \varphi \text{or}{P}_{av}=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi \\ \text{Note: Hence }\cos \varphi =\frac{R}{Z}=\text{Power factor of ac circuit} \end{gathered}$$

3.0RMS Values and Their Role

Root Mean Square (rms) Valu

It is the value of DC which would produce the same heat in a given resistance in a given time as is done by the alternating current when passed through the same resistance for the same time.

$$\begin{gathered} I_{\text{rms}}=\sqrt{\frac{{\int }_0^T{I}^2,dt}{{\int }_0^{T}dt}}\text{rms value = virtual value = Apparent value} \\ \text{rms value of}(I=I_0\sin \omega t) \\ {I}_{\text{rms}}=\sqrt{\frac{{\int }_0^T(I_0\sin \omega t{)}^2,dt}{{\int }_0^{T}dt}}=\sqrt{\frac{I_0^{,2}}{T}{\int }_0^T{\sin }^2\omega t,dt} \\ =I_0\sqrt{\frac{1}{T}{\int }_0^T\left[\frac{1-\cos 2\omega t}{2}\right]dt} \\ =I_0\sqrt{\frac{1}{T}{\left[\frac{t}{2}-\frac{\sin 2\omega t}{2\times 2\omega }\right]}_0^T}=\frac{I_0}{\sqrt{2}} \end{gathered}$$

If nothing is mentioned then values printed in a.c. circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS.

For above varieties of current $\text{rms}=\frac{\text{Peak value}}{\sqrt{2}}$

Nature of wave form	Wave-form	RMS Value	Average or mean Value
Sinusoidal		$\frac{I_0}{\sqrt{2}}=0.707I_0$	$\frac{2I_0}{\pi }=0.637I_0$
Half wave rectified		$\frac{I_0}{2}=0.5I_0$	$\frac{I_0}{\pi }=0.318I_0$
Full wave rectified		$\frac{I_0}{\sqrt{2}}=0.707I_0$	$\frac{2I_0}{\pi }=0.637I_0$

• Since AC values vary with time, Root Mean Square (RMS) values are used to calculate effective voltage and current.
• Formulas:
  $V_{\text{rms}}=\frac{V_0}{\sqrt{2}},I_{\text{rms}}=\frac{I_0}{\sqrt{2}}$
• RMS values make calculations equivalent to DC systems for power measurement.

4.0Power in AC Circuits Formula

The general formula for average power in AC circuits is:

$P=V_{\text{rms}},I_{\text{rms}}\cos \varphi$

Where:

• ($V_{rms}$): RMS voltage
• ($I_{rms}$): RMS current
• ($\cos \varphi$): Power factor (cosine of phase angle between current and voltage)

5.0Power Factor and Its Importance

• Definition:
  $\text{Power Factor}=\cos \varphi$
• Importance:
• • Determines efficiency of power transfer.
  • A power factor close to 1 indicates minimal energy loss.
  • In inductive or capacitive circuits, power factor decreases.

6.0Power in Purely Resistive, Inductive, and Capacitive Circuits

The power factor and average power are highly dependent on the type of components in the circuit.

Purely Resistive Circuit

In a purely resistive circuit, the voltage and current are in phase $(\varphi =0^{\circ })$.

• Power Factor: $\cos (\varphi =0^{\circ })=1$
• Average Power: $P_{\text{av}}=V_{\text{rms}},I_{\text{rms}}\cos \varphi$ All the apparent power is dissipated as real power.

Purely Inductive Circuit

Average Power in Inductive Circuit

$$\begin{gathered} I=I_0\cos \omega t \\ {P}_{\text{av}}=V_{\text{rms}},I_{\text{rms}}\cos 90^{\circ } \\ {P}_{\text{av}}=0 \end{gathered}$$

In a purely inductive circuit, the current lags the voltage by 90° (ϕ=+90°).

• Power Factor: cos(90°) = 0
• Average Power: $(P_{av}=V_{\text{rms}}{I}_{\text{rms}}\cos 90^{\circ })$. An ideal inductor does not dissipate any power; it simply stores and releases energy in its magnetic field.

Purely Capacitive Circuit

Average Power in Capacitive Circuit

$$\begin{gathered} I=I_0\cos \omega t \\ {P}_{av}=V_{\text{rms}}{I}_{\text{rms}}\cos 90^{\circ } \\ {P}_{av}=0 \end{gathered}$$

In a purely capacitive circuit, the current leads the voltage by 90° (ϕ=−90°).

• Power Factor: cos(−90°)=0
• Average Power: $P_{av}=V_{\text{rms}}{I}_{\text{rms}}\cos (-90^{\circ })$
• An ideal capacitor also does not dissipate any power; it stores and releases energy in its electric field.

7.0Power in RLC Series Circuits

I same for R,L and C

• Voltage and current have a phase angle ( \phi ) depending on reactance.
• Average power:
  $(P=V_{\text{rms}},I_{\text{rms}}\cos \varphi )$
• Behavior:
• • If ( R ) dominates → high power consumption.
  • If ( L ) or ( C ) dominates → reduced real power.

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• Voltage and current have a phase angle (ϕ) depending on reactance.
• Average power:
• Behavior:
• • If resistance (R) dominates → high power consumption.
  • If inductance (L) or capacitance (C) dominates → reduced real power.

8.0Different Types of Power

To fully describe power in an AC circuit, three types of power are defined. They form a right-angled triangle, known as the Power Triangle.

Apparent Power

Apparent Power (S) is the product of the RMS voltage and RMS current. It represents the total power that the source is supplying to the circuit, without considering the phase difference. It is the "potential" power.

$(S=V_{\text{rms}},I_{\text{rms}}\cos \varphi )$

The unit of apparent power is the volt-ampere (VA).

Real Power (or Average Power)

Real Power (P) is the actual power dissipated or consumed by the circuit and converted into useful work (e.g., heat, light, mechanical energy). It is the average power discussed earlier.

$P_{\text{av}}=V_{\text{rms}},I_{\text{rms}}\cos \varphi$

The unit of real power is the watt (W).

Reactive Power

Reactive Power (Q) is the power that oscillates between the source and the reactive components (inductors and capacitors). It represents the energy stored and then released by the magnetic and electric fields of these components. It does no useful work but is necessary for the operation of devices like motors and transformers.

$Q=V_{\text{rms}},I_{\text{rms}}\sin \varphi$

The unit of reactive power is the volt-ampere reactive (VAR).

Also Read: Difference Between Capacitor And Inductor

9.0The Power Triangle

The relationship between these three types of power can be visualized using a power triangle.

• The hypotenuse represents the Apparent Power (S).
• The adjacent side represents the Real Power (P).
• The opposite side represents the Reactive Power (Q).
• The angle between the real and apparent power is the phase angle ϕ.

From the Pythagorean theorem and trigonometry, we have:

$S^2=P^2+Q^{2}P=S\cos \varphi Q=S\sin \varphi$

This diagram is very helpful for solving AC power problems.

10.0Solved Examples

Illustration-1: If a direct current value a ampere is superimposed on an alternating current $I=b\sin \omega t$ flowing through a wire, what is the effective value of the resulting current in the circuit?

Solution:As current at any instant in the circuit will be,

$$\begin{gathered} I=I_{\text{DC}}+I_{\text{AC}}=a+b\sin \omega t \\ \therefore I_{\text{eff}}=\sqrt{\frac{1}{T}{\int }_0^T{I}^2,dt}=\sqrt{\frac{1}{T}{\int }_0^T(a+b\sin \omega t{)}^2,dt}=\sqrt{\frac{1}{T}{\int }_0^T\left(a^2+2ab\sin \omega t+b^2{\sin }^2\omega t\right)dt} \\ \text{But,} \\ \frac{1}{T}{\int }_0^T\sin \omega t,dt=0 \\ \text{And,}\frac{1}{T}{\int }_0^T{\sin }^2\omega t,dt=\frac{1}{2} \\ {I}_{\text{eff}}=\sqrt{a^2+\frac{1}{2}{b}^2} \end{gathered}$$

Illustration-2: If $I=2\sqrt{t}$ ampere then calculate average and rms value over t=2 to 4s. t = 2 to 4s.

Solution:

$$\begin{gathered} ⟨I⟩=\frac{{\int }_2^{4}2\sqrt{t},dt}{{\int }_2^{4}dt}=\frac{4}{3}{\left[\frac{t^{3/2}}{t}\right]}_2^4=\frac{2}{3},[8-2\sqrt{2}] \\ {I}_{\text{rms}}=\sqrt{\frac{{\int }_2^4(2\sqrt{t}{)}^2,dt}{{\int }_2^{4}dt}}=\sqrt{\frac{{\int }_2^{4}4t,dt}{2}}=\sqrt{2{\left[\frac{t^2}{2}\right]}_2^4}=2\sqrt{3} \end{gathered}$$

Illustration-3: A voltage of 10 V and frequency 103 Hz 10^3 Hz is applied to 1F (\frac{1}{\pi},\mu\text{F}) capacitor in series with a resistor of 500 500,\Omega. Find the power factor of the circuit and the power dissipated.

Solution:

$$\begin{gathered} X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 10^3\times \frac{10^{-6}}{\pi }}=500,\Omega \\ Z=\sqrt{R^2+X_C^2}=\sqrt{(500{)}^2+(500{)}^2}=500\sqrt{2},\Omega \\ \text{Power Factor }\cos \varphi =\frac{R}{Z}=\frac{500}{500\sqrt{2}}=\frac{1}{\sqrt{2}} \\ \text{Power dissipated}=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi =\frac{V_{\text{rms}}^2}{Z}\cos \varphi =\frac{(10{)}^2}{500\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{10},\text{W} \end{gathered}$$

Illustration-4: If $V=100\sin 100t\text{ volt}$ and $I=100\sin \left(100t+\frac{\pi }{3}\right)\text{ mA}$ for an A.C Circuit then find out

1. Phase difference between V and I
2. Total impedance,reactance,resistance
3. Power factor and power dissipated
4. Components contained by circuits

Solution:

$$\begin{gathered} \text{(a) Phase difference}\varphi =-\frac{\pi }{3}(I\text{ leads }V) \\ \text{(b) Total Impedance}Z=\frac{V_0}{I_0}=\frac{100}{100\times 10^{-3}}=1,\text{k}\Omega \\ \text{Now resistance}R=Z\cos 60^{\circ }=1000\times \frac{1}{2}=500,\Omega \\ \text{Reactance} \\ X=Z\sin 60^{\circ }=1000\times \frac{\sqrt{3}}{2}=\frac{500}{\sqrt{3}},\Omega \end{gathered}$$

(c)   

$$\begin{gathered} \varphi =-60^{\circ }\Rightarrow \text{Power factor}=\cos \varphi =\cos (-60^{\circ })=0.5\text{ (leading)} \\ \text{Power dissipated, }P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi =\frac{100}{\sqrt{2}}\times \frac{0.1}{\sqrt{2}}\times \frac{1}{2}=2.5,\text{W} \end{gathered}$$

$\text{Circuit must contain R as}\varphi \ne \frac{\pi }{2}andas\varphi \text{is negative. So, C must be there (L may exist but}{X}_C>X_L).$

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