A balanced Wye-connected load with (10+j 20) ohms per phase is connected to a three-phase, 400

A balanced Wye-connected load with (10+j 20) ohms per phase...

Key Concepts

Power Calculation in AC Circuits

Power in AC circuits is calculated using the concept of real power (dissipated as heat or work), reactive power (stored and released by reactive components), and apparent power. For balanced three-phase systems, the total power is typically computed by analyzing one phase and then multiplying by three, taking into account the phase voltage, current, and power factor, which reflects the phase difference between voltage and current.

AC Circuit Impedance and Phasor Analysis

Impedance in AC circuits combines resistance (the real part) and reactance (the imaginary part) into a complex quantity, which is often analyzed using phasors. Phasor analysis allows engineers to represent sinusoidal voltages and currents as rotating vectors, facilitating the calculation of circuit responses, such as current through an impedance when a particular voltage is applied.

Phase and Line Voltage Relationships

In a Wye-connected system, the phase voltage (voltage across each load) is related to the line voltage (voltage between any two phases) by a factor of ?3. Understanding this relationship is crucial for converting between line and phase quantities, which is necessary for accurate analysis and design of three-phase electrical networks.

Three-Phase Wye (Star) Connection

A Wye-connected load is one in which each phase load is connected between a phase conductor and a common neutral point. This configuration establishes a clear relationship between line voltages and phase voltages and is commonly used in power distribution systems to simplify calculations and ensure efficient load balancing.

Balanced Three-Phase Systems

This concept refers to three-phase electrical systems in which all three phases have equal magnitude voltages and are phase displaced equally by 120 degrees. In balanced systems, the loads are identical, ensuring that the currents are equal in magnitude and symmetrically displaced, which simplifies the analysis and calculation of voltages, currents, and powers.

Transcript

00:01 Given that we have a y connected load, z is equal to 10 plus z 20 om per phase.

00:08 And this load is connected to 400 volt supply.

00:12 And in this question we will find current flowing through the resistance, voltage across the resistance, power dissipated by resistance and total power consumed by this circuit.

00:27 Here i draw this circuit by using given information.

00:32 Here we have a 400 volt ac supply and this is a star connected load.

00:39 This point is neutral point and this is voltage van.

00:49 Now first we calculate van and we know van is equal to vl divided by root 3 and here i take van.

00:59 Van voltage as reference voltage so angle 0 and vl is equal to 400 volt that's why van will become equal to 400 divided by root 3 angle 0 volt here the system is balance that's why vn is equal to 0 now by looking the circuit we know ia is equal to van divided by z so now we calculate the value of ia and ia is is equal to van v n is equal to 400 divided by root 3 divided by z z is equal to 10 plus z 20 10 plus z 20 now by solving this we obtain the value of i a is equal to 10 .3279 angle negative 63 .439 degree amper now here this is the current flowing in phase a.

02:02 And in phase a, z is equal to r plus z20.

02:09 So that's why this current is also flowing through this resistance and this inductance.

02:15 Now if we want to find the voltage drop across this resistance connected in the phase a, then vra will come equal to ia into r.

02:30 And here we already obtain the value of ia that's why v r a will come equal to i a 10 .3279 angle negative 63 .4349 into r is equal to 10 of 10 by solving this we obtain the value of vr a is equals to 103 .279 angle negative 63 .439 now here the system is balanced so current ib lacks the current ia by an angle 120 degree and current ic lead ia by an angle 120 degree means i b is equal to i a angle negative 120 degree and and i c is equal to i a angle positive 120 degree now by putting the value of i a here we obtain the value of i b is equal to 10 .3279 angle 176 .5651 degree ampere and the value of ic will come equal to 10 .3329 angle 176 .5651 degree amper and the value of ic will come equal to 10 .3 29 angle 56 .56 .51g .m .m...

Please give Ace some feedback