The Power Factor Formula for Single Phase is expressed as \ ( \cos \phi = P / S \), where P is the true (active) power in watts and S is the apparent power in volt-amperes. This concept is a core part of the Class 12 Physics curriculum under the chapter on Alternating Current (NCERT Chapter 7). It is equally important for JEE Main, JEE Advanced, and NEET aspirants who encounter AC circuit problems regularly. This article covers the formula expression, derivation, a complete formula sheet, three progressively difficult solved examples, CBSE exam tips for 2025-26, common mistakes, JEE/NEET applications, and detailed FAQs.
Key Power Factor Formulas at a Glance
Quick reference for the most important single-phase power factor formulas.
- Power Factor: \( \cos \phi = \dfrac{P}{S} \)
- True (Active) Power: \( P = V I \cos \phi \)
- Apparent Power: \( S = V I \)
- Reactive Power: \( Q = V I \sin \phi \)
- Power Triangle relation: \( S^2 = P^2 + Q^2 \)
- Impedance form: \( \cos \phi = \dfrac{R}{Z} \)
- For purely resistive circuit: \( \cos \phi = 1 \)
What is the Power Factor Formula for Single Phase?
The Power Factor Formula for Single Phase gives the ratio of active (true) power consumed by a circuit to the apparent power supplied to it. In a single-phase AC circuit, voltage and current are sinusoidal. They may not be in phase with each other due to the presence of inductors or capacitors. The phase angle between voltage and current is denoted by \( \phi \). The cosine of this angle is called the power factor.
Power factor is a dimensionless quantity. Its value ranges from 0 to 1. A power factor of 1 means the circuit is purely resistive and all supplied power is consumed. A power factor of 0 means the circuit is purely reactive and no real power is consumed. This concept is introduced in NCERT Class 12 Physics, Chapter 7 — Alternating Current. It is a fundamental topic for understanding energy efficiency in electrical systems. Students appearing for CBSE board exams and competitive entrance exams must master this formula thoroughly.
Power Factor Formula for Single Phase — Expression and Variables
The primary expression for the single-phase power factor is:
\[ \cos \phi = \frac{P}{S} = \frac{\text{True Power}}{\text{Apparent Power}} \]
This can also be written in terms of circuit parameters as:
\[ \cos \phi = \frac{R}{Z} \]
And in terms of voltage and current:
\[ P = V I \cos \phi \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \cos \phi \) | Power Factor | Dimensionless (0 to 1) |
| \( P \) | True (Active) Power | Watt (W) |
| \( S \) | Apparent Power | Volt-Ampere (VA) |
| \( Q \) | Reactive Power | Volt-Ampere Reactive (VAR) |
| \( V \) | RMS Voltage | Volt (V) |
| \( I \) | RMS Current | Ampere (A) |
| \( R \) | Resistance | Ohm (Ω) |
| \( Z \) | Impedance | Ohm (Ω) |
| \( \phi \) | Phase angle between V and I | Radian or Degree |
Derivation of the Power Factor Formula
In a single-phase AC circuit, the instantaneous voltage is \( v = V_m \sin(\omega t) \) and the instantaneous current is \( i = I_m \sin(\omega t – \phi) \). The instantaneous power is \( p = vi \). Multiplying and using trigonometric identities, the average power over one full cycle is:
\[ P = \frac{V_m I_m}{2} \cos \phi = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
The apparent power is simply \( S = V_{rms} \cdot I_{rms} \). Dividing P by S gives \( \cos \phi \), which is the power factor. For an RLC series circuit, the impedance is \( Z = \sqrt{R^2 + (X_L – X_C)^2} \). The phase angle satisfies \( \cos \phi = R/Z \), which is the impedance form of the power factor formula.
Complete AC Circuit Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Power Factor (Basic) | \( \cos \phi = P / S \) | P = Active Power, S = Apparent Power | Dimensionless | Class 12, Ch 7 |
| Power Factor (Impedance) | \( \cos \phi = R / Z \) | R = Resistance, Z = Impedance | Dimensionless | Class 12, Ch 7 |
| True (Active) Power | \( P = V I \cos \phi \) | V = RMS Voltage, I = RMS Current | Watt (W) | Class 12, Ch 7 |
| Apparent Power | \( S = V I \) | V = RMS Voltage, I = RMS Current | Volt-Ampere (VA) | Class 12, Ch 7 |
| Reactive Power | \( Q = V I \sin \phi \) | V = RMS Voltage, I = RMS Current | VAR | Class 12, Ch 7 |
| Power Triangle | \( S^2 = P^2 + Q^2 \) | S = Apparent, P = Active, Q = Reactive | VA, W, VAR | Class 12, Ch 7 |
| Impedance of RLC Series | \( Z = \sqrt{R^2 + (X_L – X_C)^2} \) | R = Resistance, X_L = Inductive reactance, X_C = Capacitive reactance | Ohm (Ω) | Class 12, Ch 7 |
| Inductive Reactance | \( X_L = \omega L = 2\pi f L \) | ω = Angular frequency, L = Inductance | Ohm (Ω) | Class 12, Ch 7 |
| Capacitive Reactance | \( X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C} \) | ω = Angular frequency, C = Capacitance | Ohm (Ω) | Class 12, Ch 7 |
| Resonance Condition (Unity PF) | \( X_L = X_C \Rightarrow \cos\phi = 1 \) | X_L = Inductive reactance, X_C = Capacitive reactance | Dimensionless | Class 12, Ch 7 |
Power Factor Formula for Single Phase — Solved Examples
Example 1 (Class 11-12 Level)
Problem: A single-phase AC circuit has a resistance of 6 Ω and an inductive reactance of 8 Ω. Find the power factor of the circuit.
Given: R = 6 Ω, X_L = 8 Ω, X_C = 0 (no capacitor)
Step 1: Calculate the impedance using \( Z = \sqrt{R^2 + X_L^2} \)
\( Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \Omega \)
Step 2: Apply the power factor formula: \( \cos \phi = \dfrac{R}{Z} \)
\( \cos \phi = \dfrac{6}{10} = 0.6 \)
Step 3: The phase angle is \( \phi = \cos^{-1}(0.6) \approx 53.13^\circ \). The current lags the voltage (inductive circuit).
Answer
Power Factor = 0.6 (lagging). The circuit is inductive.
Example 2 (Class 12 / CBSE Board Level)
Problem: A single-phase AC source of 230 V (RMS) at 50 Hz is connected to a series RLC circuit with R = 10 Ω, L = 50 mH, and C = 100 μF. Calculate the power factor and the true power consumed.
Given: V = 230 V, f = 50 Hz, R = 10 Ω, L = 50 × 10⁻³ H, C = 100 × 10⁻⁶ F
Step 1: Calculate inductive reactance.
\( X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 0.05 = 15.7 \, \Omega \)
Step 2: Calculate capacitive reactance.
\( X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2 \times 3.14 \times 50 \times 100 \times 10^{-6}} = \dfrac{1}{0.03142} \approx 31.83 \, \Omega \)
Step 3: Calculate net reactance and impedance.
\( X = X_L – X_C = 15.7 – 31.83 = -16.13 \, \Omega \)
\( Z = \sqrt{R^2 + X^2} = \sqrt{10^2 + 16.13^2} = \sqrt{100 + 260.18} = \sqrt{360.18} \approx 18.98 \, \Omega \)
Step 4: Calculate power factor.
\( \cos \phi = \dfrac{R}{Z} = \dfrac{10}{18.98} \approx 0.527 \) (leading, since X_C > X_L)
Step 5: Calculate RMS current and true power.
\( I = \dfrac{V}{Z} = \dfrac{230}{18.98} \approx 12.12 \, \text{A} \)
\( P = V I \cos \phi = 230 \times 12.12 \times 0.527 \approx 1469 \, \text{W} \)
Answer
Power Factor ≈ 0.527 (leading). True Power consumed ≈ 1469 W.
Example 3 (JEE/NEET Level)
Problem: In a single-phase AC circuit, the apparent power is 500 VA and the reactive power is 300 VAR. Find the power factor and the true power. Also determine the phase angle between voltage and current.
Given: S = 500 VA, Q = 300 VAR
Step 1: Use the power triangle relation to find true power P.
\( S^2 = P^2 + Q^2 \)
\( P = \sqrt{S^2 – Q^2} = \sqrt{500^2 – 300^2} = \sqrt{250000 – 90000} = \sqrt{160000} = 400 \, \text{W} \)
Step 2: Apply the power factor formula.
\( \cos \phi = \dfrac{P}{S} = \dfrac{400}{500} = 0.8 \)
Step 3: Find the phase angle.
\( \phi = \cos^{-1}(0.8) \approx 36.87^\circ \)
Step 4: Verify using \( \sin \phi \).
\( \sin \phi = Q/S = 300/500 = 0.6 \) and \( \sin(36.87^\circ) \approx 0.6 \). Verified.
Answer
Power Factor = 0.8. True Power = 400 W. Phase angle ≈ 36.87°.
CBSE Exam Tips 2025-26
- Always state the nature of the circuit: In CBSE board exams, simply writing the numerical value of power factor is not enough. Always specify whether the circuit is resistive (PF = 1), inductive (lagging), or capacitive (leading).
- Use both forms of the formula: We recommend practising both \( \cos\phi = R/Z \) and \( \cos\phi = P/S \). The examiner may give data suited to either form. Knowing both earns full marks.
- Draw the power triangle: For 3-mark and 5-mark questions, drawing the power triangle (P, Q, S) with correct labelling fetches additional marks in CBSE 2025-26 exams.
- Remember resonance gives unity power factor: At resonance in an RLC series circuit, \( X_L = X_C \), so \( Z = R \) and \( \cos\phi = 1 \). This is a frequently tested one-liner in CBSE.
- Units matter: Active power is in Watts (W), apparent power in Volt-Ampere (VA), and reactive power in VAR. Mixing these units is a common error that costs marks.
- Revise NCERT Examples and Exercises: Our experts suggest solving all examples in NCERT Class 12 Chapter 7 before the board exam. Most CBSE questions are directly or indirectly based on these.
Common Mistakes to Avoid
- Confusing apparent power with true power: Many students substitute apparent power (S = VI) directly into power calculations. Remember, the actual power consumed is \( P = VI\cos\phi \), not VI alone. Always include the power factor term.
- Ignoring the sign of net reactance: When calculating impedance for an RLC circuit, students often forget to subtract \( X_C \) from \( X_L \). The net reactance is \( X = X_L – X_C \). Getting this wrong leads to an incorrect impedance and wrong power factor.
- Mixing peak and RMS values: The power factor formula \( P = VI\cos\phi \) uses RMS values of V and I, not peak values. Using peak values \( V_m \) and \( I_m \) without dividing by \( \sqrt{2} \) gives double the correct answer.
- Assuming power factor is always lagging: Power factor can be lagging (inductive circuits) or leading (capacitive circuits). Always check whether \( X_L > X_C \) (lagging) or \( X_C > X_L \) (leading) before stating the nature.
- Forgetting that power factor is dimensionless: Some students write units like “W/VA” for power factor in answers. Power factor is a pure ratio with no units. Writing any unit will cost marks in CBSE exams.
JEE/NEET Application of Power Factor Formula for Single Phase
In our experience, JEE aspirants encounter the Power Factor Formula for Single Phase in multiple question patterns. Understanding these patterns saves valuable time during the exam.
Pattern 1: Finding Power Factor from Circuit Elements
JEE Main frequently gives R, L, C, and frequency values. The student must calculate \( X_L \), \( X_C \), then Z, and finally \( \cos\phi = R/Z \). This is a standard 4-mark question. Practice computing these quickly without a calculator, as JEE requires mental arithmetic skills.
Pattern 2: Power Triangle Problems
JEE Advanced and NEET both test the power triangle. A typical question gives two of the three quantities (P, Q, S) and asks for the third along with the power factor. The relation \( S^2 = P^2 + Q^2 \) combined with \( \cos\phi = P/S \) solves such problems efficiently. Our experts suggest memorising the 3-4-5 and 6-8-10 Pythagorean triplets for quick mental calculation.
Pattern 3: Wattmeter and Energy Consumption
NEET occasionally asks about the reading of a wattmeter in an AC circuit. A wattmeter reads true power P, not apparent power S. If the question states a wattmeter reading and asks for apparent power or current, the student must use \( P = VI\cos\phi \) in reverse. This tests conceptual understanding of what power factor physically means. In our experience, students who understand the physical meaning — that power factor represents the fraction of apparent power actually converted to useful work — answer such questions far more accurately than those who only memorise the formula.
For further reading on AC circuit theory, refer to the official NCERT resources at ncert.nic.in.
FAQs on Power Factor Formula for Single Phase
To deepen your understanding of related concepts, explore our detailed articles on the Induced Voltage Formula, the Superposition Formula, and the Critical Velocity Formula. For a complete collection of Physics formulas aligned with NCERT, visit our Physics Formulas Hub. These resources will help you build a strong foundation for both CBSE board exams and competitive entrance tests.