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Alternating Current (AC) circuits are fundamental in electrical engineering, especially for power systems and electronics. Unlike Direct Current (DC), where voltage and current are constant, AC signals vary sinusoidally with time. Understanding the behavior of circuits under AC excitation requires special techniques.
When an AC circuit is energized, its response has two parts: the transient response and the steady state response. The transient response occurs immediately after the circuit is switched on and involves changing voltages and currents that eventually die out. The steady state response is the long-term behavior after these transients have settled, where voltages and currents vary sinusoidally at the same frequency as the source.
Steady state AC analysis focuses on this long-term sinusoidal behavior. It simplifies circuit analysis by using phasor representation and complex impedances, turning differential equations into algebraic ones. This approach is essential for solving practical problems efficiently, especially in competitive exams where time and accuracy are critical.
In this chapter, we will build from the basics of sinusoidal sources and phasors to advanced techniques like mesh and nodal analysis in AC circuits. We will also explore power calculations, resonance, and three-phase systems, all with clear examples and problem-solving strategies.
Consider a sinusoidal voltage source given by
\( v(t) = V_m \sin(\omega t + \phi) \)
where:
Analyzing circuits directly in the time domain with sinusoidal functions is cumbersome because of the trigonometric functions and derivatives involved. To simplify this, we use phasor representation.
A phasor is a complex number representing the magnitude and phase of a sinusoidal quantity, rotating in the complex plane at angular frequency \( \omega \). The sinusoidal function can be viewed as the projection of this rotating vector onto the real axis.
Thus, the voltage \( v(t) \) corresponds to a phasor \( \underline{V} \) defined as
\( \underline{V} = V_m \angle \phi \)
where \( V_m \) is the magnitude and \( \phi \) is the phase angle.
Why use phasors? Because they convert differential equations into algebraic equations by representing sinusoidal signals as complex numbers. This makes circuit analysis much simpler and faster.
In DC circuits, resistors oppose current flow by a fixed amount called resistance \( R \). In AC circuits, inductors and capacitors also oppose current, but their opposition depends on frequency and is called reactance.
To analyze AC circuits easily, we combine resistance and reactance into a single complex quantity called impedance, denoted by \( Z \). Impedance relates the phasor voltage \( \underline{V} \) and phasor current \( \underline{I} \) as
\( \underline{V} = Z \underline{I} \)
Similarly, admittance \( Y \) is the reciprocal of impedance, representing how easily current flows:
\( Y = \frac{1}{Z} \)
Let's summarize the impedance and admittance of basic circuit elements:
| Element | Impedance \( Z \) | Admittance \( Y = \frac{1}{Z} \) | Notes |
|---|---|---|---|
| Resistor (R) | \( Z_R = R \) | \( Y_R = \frac{1}{R} \) | Purely real, frequency independent |
| Inductor (L) | \( Z_L = j \omega L \) | \( Y_L = \frac{1}{j \omega L} = -j \frac{1}{\omega L} \) | Inductive reactance increases with frequency |
| Capacitor (C) | \( Z_C = \frac{1}{j \omega C} = -j \frac{1}{\omega C} \) | \( Y_C = j \omega C \) | Capacitive reactance decreases with frequency |
Important: The symbol \( j \) represents the imaginary unit, where \( j^2 = -1 \). In electrical engineering, \( j \) is used instead of \( i \) to avoid confusion with current.
Mesh and nodal analysis are systematic methods to solve circuits by writing equations based on Kirchhoff's laws. In AC steady state, these methods use phasors and complex impedances instead of time-domain voltages and currents.
The general procedure is:
graph TD A[Start: Identify AC circuit] --> B[Convert all sources to phasors] B --> C[Replace circuit elements with their impedances] C --> D[Choose mesh currents or node voltages] D --> E[Write Kirchhoff's Voltage or Current Equations using complex numbers] E --> F[Solve simultaneous equations for unknown phasors] F --> G[Convert phasors back to time domain if needed] G --> H[Analyze results: magnitude, phase, power]
Using this approach, even complex AC circuits become manageable algebraic problems. Remember to keep track of magnitudes and angles carefully.
Power in AC circuits is more complex than in DC circuits because voltage and current are sinusoidal and may not be in phase. Three types of power are important:
These are related by the power triangle, where real power is the adjacent side, reactive power the opposite side, and apparent power the hypotenuse.
The power factor (PF) is defined as
\( \text{PF} = \cos \phi = \frac{P}{|S|} \)
where \( \phi \) is the phase angle between voltage and current phasors. A power factor close to 1 means efficient power usage, while lower values indicate more reactive power.
Step 1: Identify the angular frequency \( \omega = 314 \, \text{rad/s} \) (since \( v(t) = 100 \sin(314 t) \)).
Step 2: Calculate the inductive reactance \( X_L = \omega L = 314 \times 0.2 = 62.8\,\Omega \).
Step 3: Find the total impedance \( Z = R + j X_L = 50 + j 62.8 \, \Omega \).
Step 4: Calculate the magnitude of impedance:
\( |Z| = \sqrt{50^2 + 62.8^2} = \sqrt{2500 + 3943.84} = \sqrt{6443.84} \approx 80.3\,\Omega \)
Step 5: Calculate the phase angle of impedance:
\( \theta = \tan^{-1} \left( \frac{62.8}{50} \right) = \tan^{-1}(1.256) \approx 51.6^\circ \)
Step 6: Convert source voltage to phasor form: \( \underline{V} = 100 \angle 0^\circ \) V (peak value).
Step 7: Calculate current phasor magnitude:
\( I_m = \frac{V_m}{|Z|} = \frac{100}{80.3} = 1.245\,A \)
Step 8: Current phase angle lags voltage by \( \theta \), so
\( \underline{I} = 1.245 \angle -51.6^\circ \, A \)
Answer: The current magnitude is approximately \( 1.25\,A \) and lags the voltage by \( 51.6^\circ \).
Step 1: Calculate angular frequency \( \omega = 2 \pi f = 2 \pi \times 50 = 314 \, \text{rad/s} \).
Step 2: Calculate impedances:
Step 3: Calculate admittances \( Y = \frac{1}{Z} \):
Step 4: Calculate total admittance:
\( Y_{total} = Y_R + Y_L + Y_C = 0.025 - j0.03185 + j0.0157 = 0.025 - j0.01615\,S \)
Step 5: Calculate magnitude and phase of \( Y_{total} \):
\( |Y_{total}| = \sqrt{0.025^2 + (-0.01615)^2} = \sqrt{0.000625 + 0.000261} = \sqrt{0.000886} = 0.0298\,S \)
\( \theta_Y = \tan^{-1} \left( \frac{-0.01615}{0.025} \right) = -32.9^\circ \)
Step 6: Calculate node voltage \( \underline{V} \) using source voltage \( \underline{V_s} = 120 \angle 0^\circ \, V \) and total admittance:
\( \underline{I} = Y_{total} \underline{V} \Rightarrow \underline{V} = \frac{\underline{I}}{Y_{total}} \)
Since source voltage is applied directly, node voltage is \( \underline{V} = 120 \angle 0^\circ \, V \).
Answer: The node voltage is \( 120 \angle 0^\circ \, V \). The total admittance magnitude is \( 0.0298\,S \) with phase angle \( -32.9^\circ \).
Step 1: Calculate the initial reactive power \( Q_1 \):
Given real power \( P = 50\,kW \), initial power factor \( \text{PF}_1 = 0.6 \),
\( \cos \phi_1 = 0.6 \Rightarrow \phi_1 = \cos^{-1}(0.6) = 53.13^\circ \)
Reactive power:
\( Q_1 = P \tan \phi_1 = 50 \times \tan 53.13^\circ = 50 \times 1.333 = 66.65\,kVAR \)
Step 2: Calculate reactive power after correction \( Q_2 \):
Desired power factor \( \text{PF}_2 = 0.9 \),
\( \phi_2 = \cos^{-1}(0.9) = 25.84^\circ \)
Reactive power:
\( Q_2 = P \tan \phi_2 = 50 \times \tan 25.84^\circ = 50 \times 0.484 = 24.2\,kVAR \)
Step 3: Capacitive reactive power needed:
\( Q_c = Q_1 - Q_2 = 66.65 - 24.2 = 42.45\,kVAR \)
Step 4: Calculate required capacitance \( C \):
Supply voltage \( V = 400\,V \) (line-to-line RMS), convert to phase voltage for star connection:
\( V_{ph} = \frac{400}{\sqrt{3}} = 230.94\,V \)
Capacitive reactive power formula:
\( Q_c = 3 V_{ph}^2 \omega C \Rightarrow C = \frac{Q_c}{3 V_{ph}^2 \omega} \)
Angular frequency \( \omega = 2 \pi f = 2 \pi \times 50 = 314\,rad/s \)
Calculate \( C \):
\( C = \frac{42,450}{3 \times (230.94)^2 \times 314} = \frac{42,450}{3 \times 53334 \times 314} \approx \frac{42,450}{50,250,000} = 8.45 \times 10^{-4} F = 845 \mu F \)
Answer: A capacitor of approximately \( 845 \mu F \) is required for power factor correction.
Step 1: Calculate resonant angular frequency \( \omega_0 \):
\( \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.05 \times 100 \times 10^{-6}}} = \frac{1}{\sqrt{5 \times 10^{-6}}} = \frac{1}{2.236 \times 10^{-3}} = 447.2\,rad/s \)
Step 2: Convert to frequency \( f_0 \):
\( f_0 = \frac{\omega_0}{2 \pi} = \frac{447.2}{6.283} = 71.2\,Hz \)
Step 3: Calculate impedance at resonance:
At resonance, inductive reactance equals capacitive reactance, so they cancel out:
\( X_L = X_C \Rightarrow Z = R = 10\,\Omega \)
Answer: The resonant frequency is approximately \( 71.2\,Hz \), and the impedance at resonance is purely resistive \( 10\,\Omega \).
Step 1: Identify given values:
Step 2: Use the formula for total real power in balanced three-phase system:
\( P = \sqrt{3} V_L I_L \cos \phi \)
Calculate \( P \):
\( P = 1.732 \times 415 \times 20 \times 0.8 = 1.732 \times 415 \times 16 = 1.732 \times 6640 = 11,492\,W \approx 11.5\,kW \)
Answer: The total real power consumed is approximately \( 11.5\,kW \).
When to use: To simplify calculations in steady state AC circuit problems.
When to use: When combining inductive and capacitive reactances to avoid sign errors.
When to use: To quickly estimate circuit behavior at different frequencies.
When to use: When improving power factor in industrial or residential circuits.
When to use: To understand power factor and reactive power concepts intuitively.
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