Power in AC Circuit: Definition, Formula, Derivation, Problems & Class 12 Physics Notes

Power in an AC circuit defines the rate of energy that a source requires to convert to a load for a given unit of time. This is measured in Watts. 

This AC circuit power depends on the strength of the voltage, which we call the magnitude. 

We also need to know the three types of power in AC circuits, as we cannot simply use the common formula used for DC circuits by multiplying voltage and current (as V=IR).

These 3 types of AC circuit power will have different calculations.

1. Apparent power
2. Real power
3. Reactive power

Also Read: NCERT Solutions for Class 11 & 12

We know the electrical power formula is P = V*I.
P is power, which we represent as W. V is the voltage, and I is the current. 

This is for the DC circuit, but it would not be the same for the AC circuit. One of the reasons would be that the current and voltage are not in phase. Then the same formula would be modified. 

P=Vrms×Irms×cosφ

• Vrms  is the root mean square voltage 
• Irms  is the root mean square current 
• cosϕ is the power factor 
• ϕ is the phase angle between voltage and current. 

 

1. Instantaneous Power in an AC Circuit

Assume, Voltage and current are:

$v\left(t\right)=V_{m}sin\left(\omega t\right)$

$i\left(t\right)=I_{m}sin\left(\omega t+\phi \right)$

Here,  

• $V_m$  is the maximum voltage  
• $I_m$  is the maximum current 
• ω is the angular frequency 
• ϕ is the phase difference between voltage and current 

The instantaneous power is: p(t)=v(t)×i(t) 

Substituting: $p\left(t\right)=V_m\times sin\left(\omega t\right)\times I_{m}sin\left(\omega t+\phi \right)$

• Using Trigonometric Identity: 

We know: 

sinAsinB= 1/2 [cos(A−B)−cos(A+B)]  

$p\left(t\right)=V_m{I}_m\cdot \frac{1}{2}\left[cos\left(\phi \right)-cos\left(2\omega t+\phi \right)\right]$

So, 

$p\left(t\right)=\frac{V_m{I}_m}{2}\times cos\left(\phi \right)$

• Average Power (Real Power)

Upon on full cycle, the average value of cos(2ωt+ϕ) is Zero (0).

So, 

$P_{\mathrm{avg}}=\frac{V_m{I}_m}{2}\times cos\left(\phi \right)$

Now, $V_{\mathrm{rms}}=\frac{V_m}{\sqrt{2}},I_{\mathrm{rms}}=\frac{I_m}{\sqrt{2}}$

Substituting: 

$P_{\mathrm{avg}}=V_{\mathrm{rms}}\times I_{\mathrm{rms}}\times cos\left(\phi \right)$

• Apparent and Reactive Power:

As per the definition

Apparent Power:

$S=V_{\text{rms}}\times I_{\text{rms}}\left(\text{in Volt-Amperes, VA}\right)$

Reactive Power:

$Q=V_{\mathrm{rms}}\times I_{\mathrm{rms}}\times sin\left(\phi \right)$ (in VAR)

• Power Triangle Relationship: 

Vector representation: $S^2=P^2+Q^2$

Here, 

• P is the real power 

• S is the apperant power 

• Q is the reactive power 

 

The ratio of real power (P) used to do useful work to the apparent power (S) flowing in the circuit. 

 Power Factor = Real Power / Apparent Power = cosϕ 

• Real power is measured in watts (W), which is the actual power consumed by the load. 
• Apparent Power is measured in volt-amperes (VA). It is the product of the current (I) and the RMS voltage. 
• ϕ (phi) is the phase angle between V & I. 

Related Topics: 

NCERT Class 11 notes	CBSE NCERT Class 11 Physics
Class 11 Chemistry NCERT notes	Class 11 Maths notes

Why is Power Factor Important?

Higher current is caused due to low power factor for the same power delivery, leading to: 

• Increased losses in the electrical system. 
• Over-sized equipment like conductors and transformers. 
• Improving the power factor can improve efficiency and reduce losses. 

Question 1: A circuit has $V_s=100\sqrt{2}sin(\omega t+30°)\text{and}I=2sin(\omega t+90°)$ . Find the average power.

Solution:

• Phase difference: $\phi =90°-30°=60°$
• $V_{\mathrm{rms}}=\frac{100\sqrt{2}}{\sqrt{2}}=100 V,I_{\mathrm{rms}}=\frac{2}{\sqrt{2}}=\sqrt{2} A$
• $cos60°=0.5.-P_{\text{avg}}=V_{\text{rms}}{I}_{\text{rms}}cos\phi =100\cdot \sqrt{2}\cdot 0.5=70.7\text{W}.$

Question 2: An RL circuit has $V_{\text{rms}}=200\text{V},R=10\Omega ,L=0.1\text{H},f=50\text{Hz}.$ Find $P_{\text{avg}}•|$

Solution: 

• $X_L=2\pi fL=2\pi ·50·0.1=31.4\Omega$
• $Z=\sqrt{R^2+{X_L}^2}=\sqrt{{10}^2+{31.4}^2}=32.9\Omega$
• $I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{200}{32.9}=6.08 A$
• $cos\left(\phi \right)=\frac{R}{Z}=\frac{10}{32.9}=0.304$
• $P_{\mathrm{avg}}=200\times 6.08\times 0.304=369.9 W$

Students can find the link to Class 12 Physics for all chapters. 

Sl. No	Name of Chapter
1	Chapter 1: Electric Charges and Fields
2	Chapter 2: Electrostatic Potential and Capacitance
3	Chapter 3: Current Electricity
4	Chapter 4: Moving Charges and Magnetism
5	Chapter 5: Magnetism and Matter
6	Chapter 6: Electromagnetic Induction
7	Chapter 7: Alternating Current
8	Chapter 8: Electromagnetic Waves
9	Chapter 9: Ray Optics and Optical Instruments
10	Chapter 10: Wave Optics
11	Chapter 11: Dual Nature of Radiation and Matter
12	Chapter 12: Atoms
13	Chapter 13: Nuclei
14	Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Students can find the links to the Class 12 Physics NCERT solution below. Use the solutions to prepare for the board exam.