lecture_EE1C2_lec4_power-in-AC_2025-11-30

📌 Overview

This lecture covers the analysis of power in AC circuits, distinguishing between instantaneous and average power. It introduces the conditions for maximum power transfer in the frequency domain (complex impedances) and explains the concept of RMS (Root Mean Square) or effective values for periodic signals.

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🎯 Learning Objectives

• Instantaneous Power: Understand the time-varying nature of power $p(t)=v(t)i(t)$.
• Average Power: Calculate the real power $P$ absorbed by passive elements ($R,L,C$).
• Max Power Transfer: Apply the conjugate matching theorem ($Z_L=Z_{Th}^{\ast }$) to maximize load power.
• RMS Values: Compute effective values for sinusoidal and non-sinusoidal waveforms.

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💡 Key Concepts & Definitions

• Instantaneous Power ($p(t)$): The power absorbed by an element at any specific instant in time. It consists of a constant term (average power) and a sinusoidal term at twice the frequency ($2\omega$).
• Average Power ($P$): The average of instantaneous power over one period. It represents the actual energy transfer (e.g., heat in a resistor).
  • Resistor: Absorbs average power ($P>0$).
  • Inductor/Capacitor: Average power is zero ($P=0$). They store and return energy (reactive).
• RMS (Effective) Value: The DC equivalent value that would deliver the same average power to a resistor.
• Conjugate Matching: To transfer maximum power from a source with impedance $Z_{Th}$ to a load $Z_L$, the load must be the complex conjugate of the source impedance.

➗ Formulas

$$\begin{array}{rl}\text{Instantaneous Power:}p(t)& =\underset{\text{Constant (Average)}}{\underbrace{\frac{1}{2}{V}_m{I}_m\cos ({\theta }_v-{\theta }_i)}}+\underset{\text{Oscillating (2}\omega )}{\underbrace{\frac{1}{2}{V}_m{I}_m\cos (2\omega t+{\theta }_v+{\theta }_i)}}\\ \text{Average Power:}P& =\frac{1}{T}{\int }_0^{T}p(t)dt=\frac{1}{2}{V}_m{I}_m\cos ({\theta }_v-{\theta }_i)\\ \text{RMS (General):}{X}_{rms}& =\sqrt{\frac{1}{T}{\int }_0^T{x}^2(t)dt}\\ \text{RMS (Sinusoid):}{I}_{rms}& =\frac{I_m}{\sqrt{2}},V_{rms}=\frac{V_m}{\sqrt{2}}\\ \text{Avg Power (RMS form):}P& =V_{rms}{I}_{rms}\cos ({\theta }_v-{\theta }_i)=I_{rms}^{2}R\\ \text{Max Power Condition:}{Z}_L& =Z_{Th}^{\ast }⟹R_L=R_{Th},X_L=-X_{Th}\\ \text{Max Power Value:}{P}_{max}& =\frac{|V_{Th}{|}^2}{8R_{Th}}\end{array}$$

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✍️ Notes

1. Instantaneous and Average Power

When voltage and current are sinusoidal: $v(t)=V_m\cos (\omega t+{\theta }_v)$ $i(t)=I_m\cos (\omega t+{\theta }_i)$

The Instantaneous Power is derived using the identity $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)]$:

$$p(t)=v(t)i(t)=\frac{1}{2}{V}_m{I}_m[\cos ({\theta }_v-{\theta }_i)+\cos (2\omega t+{\theta }_v+{\theta }_i)]$$

• Term 1: Constant. Depends on phase difference $\varphi ={\theta }_v-{\theta }_i$.
• Term 2: Oscillates at $2\omega$. Average is zero.

The Average Power $P$ is just the first term.

• Purely Resistive (${\theta }_v={\theta }_i$): $P=\frac{1}{2}{V}_m{I}_m$.
• Purely Reactive (L or C, ${\theta }_v-{\theta }_i=\pm 90^{\circ }$): $\cos (\pm 90^{\circ })=0⟹P=0$.

2. Maximum Average Power Transfer

We want to maximize power delivered to a load $Z_L=R_L+j{X}_L$ connected to a Thevenin equivalent circuit ($V_{Th},Z_{Th}=R_{Th}+j{X}_{Th}$).

Steps to derive condition:

1. Current magnitude: $|I|=\frac{|V_{Th}|}{\sqrt{(R_{Th}+R_L{)}^2+(X_{Th}+X_L{)}^2}}$
2. Power to load: $P=\frac{1}{2}|I{|}^2{R}_L$
3. Maximize $P$ by setting derivatives $\frac{\partial P}{\partial X_L}=0$ and $\frac{\partial P}{\partial R_L}=0$.

Result:

1. Reactance Cancellation: $X_L=-X_{Th}$
2. Resistance Matching: $R_L=R_{Th}$

Combined: $Z_L=Z_{Th}^{\ast }$ (Complex Conjugate)

Maximum Power Value: Substituting these conditions back into the power equation: $P_{max}=\frac{|V_{Th}{|}^2}{8R_{Th}}$ (Note: It is 8, not 4, because we use peak values $V_m$. If using RMS, it would be $|V_{rms}{|}^2/4R_{Th}$).

3. Effective (RMS) Value

The Effective Value of a periodic current is the DC current that delivers the same average power to a resistor. $P=R{I}_{eff}^2=\frac{1}{T}{\int }_0^{T}R{i}^2(t)dt⟹I_{eff}=\sqrt{\frac{1}{T}{\int }_0^T{i}^2(t)dt}$

• For Sinusoids: $I_{rms}=\frac{I_m}{\sqrt{2}}\approx 0.707I_m$.
• Note: Power companies supply voltage in RMS (e.g., 230V is RMS, peak is $\approx 325V$).

4. Apparent and Complex Power (Extra Context)

Although not always in the exam scope (check course guide), these define the full power picture:

• Complex Power ($S$): $S=P+jQ$ (Measured in VA).
• Apparent Power ($|S|$): $|S|=V_{rms}{I}_{rms}=\sqrt{P^2+Q^2}$.
• Real Power ($P$): $P=\text{Re}(S)$ (Measured in Watts).
• Reactive Power ($Q$): $Q=\text{Im}(S)$ (Measured in VAR).
• Power Factor: $pf=\cos ({\theta }_v-{\theta }_i)=P/|S|$.

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📝 Example Problem 1: Max Power Transfer

Problem: A circuit has a Thevenin equivalent with voltage $V_{Th}=20\angle 30^{\circ }$ V and impedance $Z_{Th}=4+j3\Omega$.

1. Determine the load impedance $Z_L$ for maximum average power transfer.
2. Calculate the maximum average power absorbed by the load.

Step-by-Step Solution:

Step 1: Identify the Thevenin Parameters

• $V_{Th}=20\angle 30^{\circ }$ V (Peak value assumed unless specified RMS)
• $Z_{Th}=4+j3\Omega$
  • $R_{Th}=4\Omega$
  • $X_{Th}=3\Omega$

Step 2: Apply Conjugate Matching Condition For maximum power, $Z_L$ must be the complex conjugate of $Z_{Th}$. $Z_L=Z_{Th}^{\ast }=R_{Th}-j{X}_{Th}$ $Z_L=4-j3\Omega$ (Ideally, this means a $4\Omega$ resistor in series with a capacitor having $-j3\Omega$ reactance).

Step 3: Calculate Maximum Average Power Use the derived formula for $P_{max}$. $P_{max}=\frac{|V_{Th}{|}^2}{8R_{Th}}$

• $|V_{Th}|=20$ V
• $R_{Th}=4\Omega$

$P_{max}=\frac{20^2}{8\cdot 4}=\frac{400}{32}=12.5\text{ W}$

Verification (Optional but good practice): Calculate current $I$: Total Impedance $Z_{tot}=Z_{Th}+Z_L=(4+j3)+(4-j3)=8\Omega$ (Purely resistive!). Current $I=\frac{V_{Th}}{Z_{tot}}=\frac{20\angle 30^{\circ }}{8}=2.5\angle 30^{\circ }$ A. Power $P=\frac{1}{2}|I{|}^2{R}_L=\frac{1}{2}(2.5{)}^2(4)=\frac{1}{2}(6.25)(4)=12.5\text{ W}$.

Rare Case Note: If the load is constrained to be purely resistive ($Z_L=R_L$), we cannot simply set $Z_L=Z_{Th}^{\ast }$. Instead, we match the magnitude: $R_L=|Z_{Th}|=\sqrt{R_{Th}^2+X_{Th}^2}$. The power will be lower than the true maximum.

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📝 Example Problem 2: Average Power Calculation

Problem: Calculate the average power absorbed by an impedance if the voltage is $v(t)=100\cos (\omega t+15^{\circ })$ V and current is $i(t)=4\sin (\omega t-15^{\circ })$ A.

Step-by-Step Solution:

Step 1: Ensure Consistent Functions (Cos/Cos) The current is given in sine. Convert to cosine: $\sin (x)=\cos (x-90^{\circ })$. $i(t)=4\cos (\omega t-15^{\circ }-90^{\circ })=4\cos (\omega t-105^{\circ })\text{ A}$

Step 2: Identify Parameters

• $V_m=100$ V, ${\theta }_v=15^{\circ }$
• $I_m=4$ A, ${\theta }_i=-105^{\circ }$

Step 3: Apply Average Power Formula $P=\frac{1}{2}{V}_m{I}_m\cos ({\theta }_v-{\theta }_i)$ $P=\frac{1}{2}(100)(4)\cos (15^{\circ }-(-105^{\circ }))$ $P=200\cos (120^{\circ })$ $P=200(-0.5)=-100\text{ W}$

Step 4: Interpret Result The power is negative, meaning the element is delivering 100 W to the rest of the circuit (active source), rather than absorbing it.

⚠️ Warning: Always check if values are Peak ($V_m$) or RMS ($V_{rms}$).

• If Peak: $P=\frac{1}{2}{V}_m{I}_m\cos (\theta )$
• If RMS: $P=V_{rms}{I}_{rms}\cos (\theta )$

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📝 Example Problem 3: RMS Application

Problem: A sinusoidal voltage with a maximum amplitude of $V_m=625$ V is applied to a $50\Omega$ resistor. Calculate the average power delivered.

Step-by-Step Solution:

Step 1: Determine the calculation method Since it’s a resistor, we can use $P=\frac{V_{rms}^2}{R}$ or $P=\frac{1}{2}\frac{V_m^2}{R}$.

Step 2: Calculate via RMS (Standard practice) $V_{rms}=\frac{V_m}{\sqrt{2}}=\frac{625}{\sqrt{2}}\approx 441.94\text{ V}$ $P=\frac{V_{rms}^2}{R}=\frac{(441.94{)}^2}{50}$

Step 3: Calculate via Peak Formula (Verification) $P=\frac{1}{2}\frac{V_m^2}{R}=\frac{1}{2}\frac{625^2}{50}$ $P=\frac{1}{2}\frac{390625}{50}=\frac{7812.5}{2}=3906.25\text{ W}$

Result: The average power is 3906.25 W (or approx 3.9 kW).

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🔗 Resources

• Presentation:

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❓ Post lecture

• Key Takeaway: Reactive elements (L, C) do not consume average power; they only exchange it. To maximize real power transfer, you must cancel out the source reactance with an opposite load reactance.

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📖 Homework

• Practice calculating RMS for non-sinusoidal waves (e.g., triangle or square waves).
• Solve problems where $Z_L$ is restricted (e.g., only resistive).