AC fundamentals and single-phase circuits notes — Unit 2

Alternating Current (AC) is an electric current that periodically reverses its direction and changes its magnitude continuously with time. The most standard and useful form is a sinusoidal wave.

⇒Why AC is preferred over DC:

• AC voltage can be easily stepped up or down using transformers, minimizing transmission losses over long distances.
• AC motors are much simpler, robust, and cheaper than DC motors.
• High voltage AC generation is easier than DC generation.

A sinusoidal voltage is generated by rotating a rectangular coil in a uniform magnetic field at a constant angular velocity ($\omega$), obeying Faraday's Law of Induction.

v(t) = V_m sin(ωt)

Where:
v(t) = Instantaneous voltage
V_m = Maximum (Peak) voltage
ω = Angular frequency in rad/s (ω = 2πf)

The same applies for current: $i(t) = I_m \sin(\omega t)$.

• Cycle: One complete set of positive and negative values of an alternating quantity.
• Time Period ($T$): The time taken to complete one full cycle (measured in seconds).
• Frequency ($f$): The number of cycles completed in one second (measured in Hertz, Hz). $f = 1/T$. In India, grid frequency is 50 Hz.
• Amplitude (Peak Value): The maximum positive or negative value reached in a cycle.

⇒Average Value

The average value of a full sine wave is zero (positive half cancels negative half). So we calculate the average over a half cycle.

V_avg = (1/π) ∫₀^π V_m sinθ dθ
V_avg = (2/π) V_m ≈ 0.637 V_m

⇒Root Mean Square (RMS) Value

The RMS value of an AC current is that steady DC current which produces the same heating effect as the AC current in the same resistor over the same time.

To find the RMS value, we square the function, find the mean over a full cycle, and take the square root.

V_rms = √ [ (1/2π) ∫₀^2π (V_m sinθ)² dθ ]
V_rms = V_m / √2 ≈ 0.707 V_m

These factors tell us about the shape of the waveform.

⇒Form Factor ($K_f$)

Ratio of RMS value to Average value.

K_f = RMS / Average = (V_m/√2) / (2V_m/π) = 1.11
(For a pure sine wave)

⇒Crest (Peak) Factor ($K_p$)

Ratio of Peak value to RMS value.

K_p = Peak / RMS = V_m / (V_m/√2) = 1.414 (√2)

If two AC waves have the same frequency, they may not peak at the same time. This time difference is measured in degrees or radians and called the Phase Difference ($\phi$).

• In-Phase: Both waves reach zero and peak at the exact same time ($\phi = 0$).
• Leading: Wave A is said to lead Wave B if A reaches its peak before B.
• Lagging: Wave B lags Wave A.

v = V_m sin(ωt)
i = I_m sin(ωt - φ)  --> Current LAGS voltage by φ
i = I_m sin(ωt + φ)  --> Current LEADS voltage by φ

Instead of dealing with trigonometric sine waves, we represent AC quantities as Phasors—rotating vectors in the complex plane.

A phasor has a magnitude equal to the RMS value of the wave, and an angle equal to its phase angle. It rotates counter-clockwise at angular velocity $\omega$.

Polar Form: V = |V| ∠φ
Rectangular Form: V = V_real + j(V_imag)
V = |V| (cosφ + j sinφ)

This converts hard calculus/trigonometry into simple complex number algebra.

Apply $v = V_m \sin(\omega t)$ to a pure resistor $R$.

By Ohm's law, $i = v/R = (V_m/R) \sin(\omega t) = I_m \sin(\omega t)$.

• Phase Relation: Voltage and Current are exactly in-phase ($\phi = 0$).
• Power: Average power $P = V_{rms} I_{rms} \cos(0) = VI$.

The phasor diagram shows both the V vector and I vector pointing in the same direction.

Apply $v = V_m \sin(\omega t)$ to a pure inductor $L$.

Using $v = L(di/dt)$, integration gives $i = I_m \sin(\omega t - 90^\circ)$.

• Phase Relation: Current LAGS the voltage by exactly 90°. (ELI: Voltage E leads Current I in Inductor L).
• Power: Average power $P = V I \cos(90^\circ) = 0$. A pure inductor consumes zero average power!

⇒Inductive Reactance ($X_L$)

The opposition offered by the inductor. $X_L = \omega L = 2\pi f L$ (measured in Ohms).

Apply $v = V_m \sin(\omega t)$ to a pure capacitor $C$.

Using $i = C(dv/dt)$, differentiation gives $i = I_m \sin(\omega t + 90^\circ)$.

• Phase Relation: Current LEADS the voltage by exactly 90°. (ICE: Current I leads Voltage E in Capacitor C).
• Power: Average power $P = V I \cos(-90^\circ) = 0$. A pure capacitor consumes zero average power.

⇒Capacitive Reactance ($X_C$)

The opposition offered by the capacitor. $X_C = 1 / (\omega C) = 1 / (2\pi f C)$ (measured in Ohms).

In real circuits containing R, L, and C, the total opposition to AC flow is called Impedance (Z), measured in Ohms.

Impedance is a complex number: $Z = R + jX$, where $R$ is resistance and $X$ is net reactance ($X_L - X_C$).

Magnitude: |Z| = √(R² + X²)
Phase angle: φ = tan⁻¹(X / R)

Ohm's Law for AC circuits is written using phasors: $\mathbf{V} = \mathbf{I} \cdot \mathbf{Z}$.

A resistor $R$ and inductor $L$ in series.

• Impedance $Z = R + jX_L$.
• Magnitude $|Z| = \sqrt{R^2 + X_L^2}$.
• Current $I$ lags applied Voltage $V$ by angle $\phi = \tan^{-1}(X_L/R)$. (Angle is between 0° and 90°).

⇒Impedance Triangle

A right-angled triangle with base $R$, perpendicular $X_L$, and hypotenuse $Z$. The angle between $R$ and $Z$ is the phase angle $\phi$.

All three elements (R, L, C) in series.

Net Impedance $Z = R + j(X_L - X_C)$.

• Case 1: $X_L > X_C$: The circuit behaves inductively. Current lags voltage. $\phi$ is positive.
• Case 2: $X_C > X_L$: The circuit behaves capacitively. Current leads voltage. $\phi$ is negative.
• Case 3: $X_L = X_C$: The reactive parts cancel out! The circuit behaves as a pure resistance. This is called Resonance.

Resonance occurs in an RLC circuit when Inductive Reactance equals Capacitive Reactance ($X_L = X_C$).

ωL = 1 / (ωC)
ω² = 1 / (LC)
f_r = 1 / (2π √(LC))

⇒Properties at Resonance

• Impedance is minimum and purely real ($Z = R$).
• Current is maximum ($I = V/R$).
• Power factor is exactly unity (1).
• Voltage across L and C can be much higher than the supply voltage (Voltage Magnification).

Quality Factor (Q-factor) is a measure of the "sharpness" of resonance. It is defined as the ratio of voltage across L or C to the applied voltage at resonance.

Q = (1/R) * √(L/C)

A highly selective circuit (like a radio tuner) has a low resistance $R$ and thus a very high Q-factor, resulting in a sharp peak at resonance.

Bandwidth is the range of frequencies over which the current is at least $1/\sqrt{2}$ (or 70.7%) of its maximum resonant value. $BW = f_r / Q$.

In AC circuits, power is split into three types, forming a right-angled Power Triangle:

• Apparent Power (S): Total power supplied. $S = VI$. Unit: Volt-Amperes (VA). (The Hypotenuse).
• Active/Real Power (P): Actual power consumed by resistors and converted to useful work. $P = VI \cos\phi$. Unit: Watts (W). (The Base).
• Reactive Power (Q): Power sloshing back and forth between source and L/C fields. Does no useful work. $Q = VI \sin\phi$. Unit: Volt-Amperes Reactive (VAR). (The Perpendicular).

S² = P² + Q²
S = P + jQ (Complex Power)

Power Factor is the cosine of the phase angle between voltage and current. It is also the ratio of Active Power to Apparent Power ($P/S$).

It indicates how efficiently electrical power is being converted into useful work output.

• Ideal Power Factor = 1 (Purely resistive circuit). All supplied power does work.
• Worst Power Factor = 0 (Pure L or C). Current flows, but zero work is done.

A low power factor means the power company has to push a massive current (Apparent power) just to deliver a small amount of Real power, causing huge $I^2R$ heating losses in transmission lines.

Most industrial loads (induction motors) are highly inductive, causing a lagging power factor and wasting grid capacity.

We can "correct" (improve) the power factor closer to 1 by connecting Capacitor Banks in parallel with the inductive load.

The capacitor draws a leading reactive current that exactly cancels out the lagging reactive current of the inductor. The source now only has to supply the active current, vastly reducing transmission losses.

⇒Unit 2 Mastery

• Derive the RMS and Average values of a pure sine wave.
• Draw the phasor diagram and impedance triangle for an R-L series circuit.
• State the condition for series resonance and derive the resonant frequency formula.
• Define Active, Reactive, and Apparent power, and draw the Power Triangle.
• Why is a low power factor bad, and how is it corrected?