Essential AC Circuit Analysis & Formulas
1. Understanding RMS Values in AC Circuits
The RMS (Root Mean Square) current value of an alternating current is defined as the equivalent direct current that produces the same heating effect. This equivalence allows the use of DC calculation methods for alternating current circuits.
Since instantaneous values of current and voltage are variable, RMS values are used to represent these magnitudes in a way that is useful over time. RMS values are practical for measurements in AC circuits and are commonly used in calculations. For example, when referring to a domestic electrical voltage of 200V, we are referring to its effective (RMS) value.
2. Analyzing AC EMF Expressions
The instantaneous value of an AC EMF is given by the expression E(t) = 200 × sin(314t). Calculate:
Given E(t) = Emax · sin(ωt)
a. Maximum EMF
Emax = 200 V
b. Angular Frequency (Pulsation)
ω = 314 rad/s
c. Period
T = 2π / ω = 2π / 314 ≈ 0.02 s
d. Frequency
f = 1 / T = 50 Hz
3. AC Current Calculations: Frequency & Intensity
An AC current has a frequency of 50 Hz and a maximum intensity of 10 A. Calculate:
a. Angular Frequency (Pulsation)
ω = 2πf = 2π · 50 = 314 rad/s
b. Period
T = 1 / f = 1 / 50 = 0.02 s
c. Instantaneous Value
I(t) = Imax · sin(ωt) = 10 · sin(314t)
d. Average Value (Half-Cycle)
The average value of a complete sine wave cycle is zero.
The average value over a half-cycle of a sine wave is:
Iavg = (2 / π) · Imax = (2 / π) · 10 = 6.37 A
4. Sinusoidal AC Current Analysis
Given the sinusoidal AC current I(t) = 5 · sin(314t), calculate:
a. Frequency
f = ω / (2π) = 314 / (2π) = 50 Hz
b. Period
T = 1 / f = 1 / 50 = 0.02 s
c. Maximum and Effective Intensities
Since I(t) = Imax · sin(ωt)
Imax = 5 A
Irms = Imax / √2 = 5 / √2 ≈ 3.54 A
d. Instantaneous Value at t = 0.005 s and at an Angle of 150°
For t = 0.005 s:
I(t) = Imax · sin(ωt) = 5 · sin(314 · 0.005) = 5 · sin(1.57 rad) = 5 · sin(90°) = 5 A
For an angle of 150°:
I(t) = Imax · sin(θ) = 5 · sin(150°) = 2.5 A
5. AC Voltage Calculations: Frequency & RMS Value
The frequency and RMS voltage measurements for an AC circuit are, respectively, f = 50 Hz and Vrms = 250 V. Calculate:
a. Maximum Voltage Value
Vmax = Vrms · √2 = 250 · √2 ≈ 354 V
b. Angular Frequency (Pulsation)
ω = 2πf = 2π · 50 = 314 rad/s
c. Instantaneous Voltage at 0.005 s
V(t) = Vmax · sin(ωt) = 354 · sin(314 · 0.005) = 354 · sin(1.57 rad) = 354 · sin(90°) = 354 V
6. Phase Angle in Inductive & Capacitive Circuits
What is the sign of the angle between the voltage and current in an inductive and capacitive circuit?
The phase angle in an inductive circuit is positive, and in a capacitive circuit, it is negative.
In an inductive circuit, the current lags behind the voltage, meaning the voltage phase is always greater than the current phase. Conversely, in a capacitive circuit, the current leads the voltage, meaning the voltage phase is always less than the current phase.
7. Current Behavior in Pure Capacitive Circuits
Using the sinusoidal representation of voltage and current, explain how current behaves in a pure capacitive circuit during charging and discharging of the capacitor.
In the first and third quarter-cycles of the capacitor’s charging period, the current decreases as the capacitor voltage increases.
In the second and fourth quarter-cycles, the capacitor discharges. The current trend is increasing as the capacitor voltage decreases.
8. Resistive AC Circuit Analysis
An electrical resistance of 1000 Ω is connected to a sinusoidal alternating voltage of 220 V and 50 Hz. Calculate the RMS, maximum, and instantaneous values. Represent the phasor diagram.
Irms = Vrms / R = 220 V / 1000 Ω = 0.22 A
Imax = Irms · √2 = 0.22 · √2 ≈ 0.311 A
i(t) = Imax · sin(ωt) = 0.311 · sin(2π · 50 · t) = 0.311 · sin(314t)
Phasor Diagram
Ivector = Vvector / ZR_vector = 220∠0° / 1000∠0° = 0.22∠0° A
9. Inductive AC Circuit Calculations
A 100 mH coil is connected to a voltage of 125 V and 70 Hz. Determine:
a. Inductive Reactance
XL = 2πfL = 2π · 70 Hz · 100 × 10-3 H ≈ 44 Ω
b. Effective and Maximum Intensities
Irms = Vrms / XL = 125 V / 44 Ω ≈ 2.84 A
Imax = Irms · √2 = 2.84 · √2 ≈ 4.02 A
c. Expression for Instantaneous Current
i(t) = Imax · sin(ωt – π/2) = 4.02 · sin(2π · 70 · t – π/2) = 4.02 · sin(440t – π/2)
d. Phasor Diagram
Ivector = Vvector / ZL_vector = 125∠0° / 44∠90° = 2.84∠-90° A
10. Capacitive AC Circuit Calculations
A 20 μF capacitor is connected to a sinusoidal AC voltage of 380 V and 50 Hz. Determine:
a. Capacitive Reactance
XC = 1 / (2πfC) = 1 / (2π · 50 Hz · 20 × 10-6 F) ≈ 159 Ω
b. Effective and Maximum Intensities
Irms = Vrms / XC = 380 V / 159 Ω ≈ 2.39 A
Imax = Irms · √2 = 2.39 · √2 ≈ 3.38 A
c. Instantaneous Current
i(t) = Imax · sin(ωt + π/2) = 3.38 · sin(2π · 50 · t + π/2) = 3.38 · sin(314t + π/2)
d. Phasor Diagram
Ivector = Vvector / ZC_vector = 380∠0° / 159∠-90° = 2.39∠90° A