Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Power in AC circuits
Download the Lessonotes Mobile Nigeria 2025 app for faster lesson access on Android and iPhone.
Power in AC circuits
Download the Lessonotes Mobile Nigeria 2025 app for faster lesson access on Android and iPhone.
Subject: Basic Electronics
Class: Senior Secondary 2
Term: 2nd Term
Week: 1
Theme: Electronic Components And Circuits
This page supports the lesson note with a companion video and a short classroom-ready summary.
For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.
Explain power and power triangle. Explain power factor and power factor correction. State advantages and disadvantages of power factor correction. Calculate power factor in a given AC circuit. Explain Q-factor and bandwidth.
In AC circuits, power behaves differently compared to DC circuits due to the presence of reactive components (inductors and capacitors) which cause phase shifts between voltage and current.
Three types of power are considered: Real Power (Active Power or True Power), P: Definition: This is the actual power consumed by the resistive components of a circuit and is responsible for doing useful work. It is the power that performs mechanical work, produces heat, or generates light.
Unit: Watts (W).
Formula: $P = V_{rms} \cdot I_{rms} \cdot \cos(\phi)$ Where: $V_{rms}$ is the RMS (Root Mean Square) voltage. $I_{rms}$ is the RMS current. $\phi$ (phi) is the phase angle between the voltage and current. $\cos(\phi)$ is the power factor.
Example: The power consumed by a heater, a light bulb, or the useful output of an electric motor. Reactive Power, Q: Definition: This is the power that flows back and forth between the source and the reactive components (inductors and capacitors) of the circuit. It is not consumed and does not perform useful work, but it is necessary for the operation of magnetic devices (like motors, transformers) and for charging/discharging capacitors.
Unit: Volt-Ampere Reactive (VAR).
Formula: $Q = V_{rms} \cdot I_{rms} \cdot \sin(\phi)$ Nature: Inductive loads (motors, transformers) consume lagging reactive power (current lags voltage), while capacitive loads (capacitors) generate leading reactive power (current leads voltage). Apparent Power, S: Definition: This is the total power supplied by the source, which is the product of the RMS voltage and the RMS current, irrespective of the phase angle. It represents the total capacity of the power supply system.
Unit: Volt-Ampere (VA).
Formula: $S = V_{rms} \cdot I_{rms}$ Relationship to P and Q: Apparent power is the vector sum of real power and reactive power. The relationship between real power (P), reactive power (Q), and apparent power (S) can be visualized using a right-angled triangle called the Power Triangle. This analogy is crucial for understanding power factor.
Hypotenuse: Apparent Power (S)
Adjacent Side: Real Power (P)
Opposite Side: Reactive Power (Q)
Angle between S and P: The phase angle $\phi$ (phi).
Diagram: ``` S (Apparent Power, VA) /| / | / | Q (Reactive Power, VAR) / | /__φ_| P (Real Power, W) ``` Pythagorean Relationship: $S^2 = P^2 + Q^2$ Definition: Power factor is a measure of how effectively electrical power is being converted into useful work output. It is defined as the cosine of the phase angle ($\phi$) between the voltage and current waveforms in an AC circuit. It is also the ratio of real power to apparent power.
Formula: $PF = \cos(\phi)$ $PF = P/S$ (Real Power / Apparent Power)
Range: Power factor can range from 0 to 1 (or 0% to 100%).
Interpretation: PF = 1 (Unity Power Factor): This occurs when the voltage and current are perfectly in phase ($\phi = 0^\circ$). All the apparent power is real power, meaning maximum efficiency. This is ideal but rarely achieved in practical inductive circuits. *PF 0^\circ$). This is the most common scenario in Nigerian industries and homes. PF < 1 (Leading Power Factor): Occurs in circuits with capacitive loads where the current leads the voltage ($\phi < 0^\circ$). This can happen if too many capacitors are used for power factor correction.
Importance: A low power factor indicates that a larger apparent power (S) is being supplied for the same useful real power (P).
This means: More current flows through the lines for the same amount of useful power. Increased losses in transmission and distribution lines ($I^2R$ losses). Larger capacity equipment (transformers, generators, cables) is needed. Higher electricity bills for consumers (especially industrial) due to penalties from power companies (e.g., DISCOs) for excessive reactive power.
Worked Example 1: Calculating Power Parameters An industrial motor in a Nigerian factory operates on a 400V, 50Hz AC supply. It draws a current of 25A, and the current lags the voltage by $36.87^\circ$.
Calculate: a) Apparent Power (S) b) Real Power (P) c) Reactive Power (Q) d)
Power Factor (PF)
Solution: Given: $V_{rms} = 400V$, $I_{rms} = 25A$, $\phi = 36.87^\circ$ a)
Apparent Power (S): $S = V_{rms} \cdot I_{rms}$ $S = 400V \cdot 25A$ $S = 10,000 VA$ or $10 kVA$ b)
Real Power (P): $P = V_{rms} \cdot I_{rms} \cdot \cos(\phi)$ $P = 400V \cdot 25A \cdot \cos(36.87^\circ)$ $P = 10,000 \cdot 0.8$ $P = 8,000 W$ or $8 kW$ c)
Reactive Power (Q): $Q = V_{rms} \cdot I_{rms} \cdot \sin(\phi)$ $Q = 400V \cdot 25A \cdot \sin(36.87^\circ)$ $Q = 10,000 \cdot 0.6$ $Q = 6,000 VAR$ or $6 kVAR$ d)
Power Factor (PF): $PF = \cos(\phi)$ $PF = \cos(36.87^\circ)$ $PF = 0.8$ (lagging) Alternatively, $PF = P/S = 8000W / 10000VA = 0.8$ Explanation: Power factor correction is the process of improving the power factor of an AC circuit, typically by bringing it closer to unity (1). This is usually achieved by adding capacitors in parallel with inductive loads.
Mechanism: Inductive loads consume lagging reactive power. Capacitors, on the other hand, supply leading reactive power. By connecting capacitors in parallel, the reactive power supplied by the capacitors cancels out a portion of the lagging reactive power consumed by the inductive load, thereby reducing the total reactive power drawn from the main supply. This reduces the phase angle ($\phi$) between the total current and voltage, increasing the power factor.
Worked Example 2: Power Factor Correction The industrial motor from Example 1 has a real power of 8 kW and a reactive power of 6 kVAR (lagging) at 400V, 50Hz. The factory wants to improve the power factor to 0.95 lagging. Calculate the capacitance value (C) required to achieve this.
Solution: Given: $P = 8000W$, $Q_{initial} = 6000 VAR$, $V = 400V$, $f = 50Hz$. Desired $PF_{new} = 0.95$ (lagging). Find the new phase angle ($\phi_{new}$): $\cos(\phi_{new}) = PF_{new} = 0.95$ $\phi_{new} = \arccos(0.95) \approx 18.19^\circ$ Calculate the new apparent power ($S_{new}$): $P = S_{new} \cdot \cos(\phi_{new})$ $S_{new} = P / \cos(\phi_{new}) = 8000W / 0.95 \approx 8421.05 VA$ Calculate the new reactive power ($Q_{new}$): $Q_{new} = S_{new} \cdot \sin(\phi_{new})$ $Q_{new} = 8421.05 VA \cdot \sin(18.19^\circ)$ $Q_{new} = 8421.05 VA \cdot 0.3121$ $Q_{new} \approx 2626.4 VAR$ (lagging) Calculate the reactive power supplied by the capacitor ($Q_C$): The capacitor needs to supply reactive power equal to the difference between the initial reactive power and the new desired reactive power. $Q_C = Q_{initial} - Q_{new}$ $Q_C = 6000 VAR - 2626.4 VAR$ $Q_C \approx 3373.6 VAR$ Calculate the capacitance (C): The formula for reactive power of a capacitor is $Q_C = V^2 / X_C$, where $X_C = 1 / (2 \pi f C)$. So, $Q_C = V^2 \cdot 2 \pi f C$ $C = Q_C / (V^2 \cdot 2 \pi f)$ $C = 3373.6 VAR / ((400V)^2 \cdot 2 \cdot \pi \cdot 50Hz)$ $C = 3373.6 / (160000 \cdot 314.159)$ $C = 3373.6 / 50265440$ $C \approx 6.71 \times 10^{-5} F$ $C \approx 67.1 \mu F$
Electricity Billing and Penalties in Industries (Economic Impact): Many industries and large commercial consumers in Nigeria are billed based on both active power (kWh) and reactive power (kVARh) or peak demand (kVA). A low power factor means the industry draws more apparent power for the same useful work, leading to higher kVA demand charges or specific low power factor penalties from electricity distribution companies (DISCOs). Implementing power factor correction directly reduces these charges, saving significant operational costs for Nigerian businesses like textile mills in Kaduna, cement factories in Ewekoro, or food processing plants in Ibadan. National Grid Stability and Efficiency (Community/Environment): The national grid (managed by TCN and DISCOs) operates more efficiently when the overall power factor is high. Low power factor on a large scale leads to increased current in transmission lines, causing higher $I^2R$ losses (heat waste), greater voltage drops, and reduced grid capacity. By understanding and correcting power factor, students can appreciate how efficient power usage at the consumer end contributes to a more stable, reliable, and less wasteful electricity supply nationwide, potentially reducing the frequency of power outages or the need for expensive grid infrastructure upgrades. This also has environmental benefits by reducing energy waste. Radio and Communication Systems (Technological/Social): The concepts of Q-factor and bandwidth are crucial in the design and operation of radio receivers, mobile phones, and other communication devices widely used in Nigeria. A high Q-factor in a radio receiver allows it to sharply tune into a specific radio station (e.g., Radio Nigeria, Wazobia FM) while rejecting interference from adjacent stations. The bandwidth determines how much information (audio, video, data) can be transmitted over a particular channel. Understanding these principles is vital for electronics technicians involved in the repair, maintenance, and even local design of communication equipment.