Three Phase AC Circuits : Problem 1 - Three Phase Circuits - Basic Electrical Engineering

Summary

In this video, we solve a problem related to a three-phase Delta-connected circuit. The problem involves a load with a 50 mH inductor in series with a parallel combination of a 50 Ohm resistor and a 50 μF capacitor, connected to a 550V AC supply. The video guides through calculating phase impedance (Zph), phase current (Iph), line current, active power, power factor, reactive power, and KVA rating. The solution breaks down key formulas, including Ohm’s law and the relationships between line and phase values in a Delta circuit. Practical tips for handling similar problems are also discussed.

Takeaways

• 🔧 Each phase of a Delta-connected load consists of a 50 mH inductor in series with a parallel combination of a 50 Ohm resistor and a 50 microfarad capacitor.
• ⚡ The given AC supply is a three-phase, 550 volts, 800 radians per second system.
• 📐 The impedance (Zph) of each phase must be calculated using the inductive reactance (XL) and capacitive reactance (XC).
• 📏 Inductive reactance (XL) is calculated as 40 Ohms, and capacitive reactance (XC) is calculated as 25 Ohms.
• 🔄 Zph is a combination of the inductive reactance in series with the parallel combination of the resistor and capacitive reactance.
• 💡 The calculated impedance angle (Zph angle) determines the power factor angle and the nature of the load (inductive or capacitive).
• 🔋 Phase voltage is equal to line voltage for a Delta-connected load, which is given as 550 volts.
• 🔢 Phase current (Iph) is obtained using Ohm's law, and line current (IL) is calculated as root 3 times the phase current.
• 📊 The active power (P), reactive power (Q), and apparent power (S) are calculated using respective formulas.
• 🏷️ The final results: Power drawn is 1815.31 kW, power factor is 0.4473 (lagging), reactive power is 36.2984 kVAR, and KVA rating is 40.5846 KVA.

Q & A

What type of circuit is being analyzed in the problem?

The problem involves a three-phase Delta-connected circuit with each phase containing a 50 mH inductor in series with a parallel combination of a 50 Ohm resistor and a 50 microfarad capacitor.

What are the key components in each phase of the Delta-connected load?

Each phase consists of a 50 mH inductor, a 50 Ohm resistor, and a 50 microfarad capacitor arranged in a series-parallel configuration.

How is the inductive reactance (XL) calculated?

The inductive reactance (XL) is calculated using the formula XL = Omega * L. Given Omega is 800 rad/s and L is 50 mH, XL is 40 Ohms.

How is the capacitive reactance (XC) calculated?

The capacitive reactance (XC) is calculated using the formula XC = 1 / (Omega * C). With Omega = 800 rad/s and C = 50 microfarads, XC equals 25 Ohms.

What is the significance of considering the reactances as complex numbers?

Reactances are considered complex numbers because inductive reactance (XL) is positive imaginary (j40 Ohms) and capacitive reactance (XC) is negative imaginary (-j25 Ohms). This is important for accurate calculation of impedance in AC circuits.

How is the phase impedance (Zph) calculated?

Zph is the series combination of inductive reactance (j40 Ohms) and the parallel combination of the resistor (50 Ohms) with capacitive reactance (-j25 Ohms). Using complex impedance formulas, Zph is 22.36 Ohms at an angle of 63.43 degrees.

What is the relationship between phase voltage and line voltage in a Delta-connected load?

In a Delta-connected load, the phase voltage is the same as the line voltage. Therefore, the phase voltage is 550 V, equal to the line voltage.

How is the phase current (Iph) determined?

Phase current (Iph) is determined using Ohm's law: Iph = Vph / Zph. Substituting Vph = 550 V and Zph = 22.36 Ohms, Iph is calculated to be 24.6 A.

How is the line current (IL) calculated for a Delta-connected load?

For a Delta-connected load, line current (IL) is calculated using the formula IL = sqrt(3) * Iph. Given Iph = 24.6 A, IL is 42.6 A.

What is the active power (P) drawn by the load?

The active power (P) is calculated using the formula P = 3 * Vph * Iph * cos(ϕ). Substituting the values, the power drawn is 1815.31 kW.

How is the reactive power (Q) calculated?

Reactive power (Q) is calculated using the formula Q = 3 * Vph * Iph * sin(ϕ). With the given values, Q is 36.3 kVAR.

How is the apparent power (S) determined?

The apparent power (S) is determined using the formula S = 3 * Vph * Iph. Substituting the values, S is 4058.46 kVA.