Three equal star-connected inductors consume 8 kW power at 0.8 power factor when connected to 4

Three equal star-connected inductors consume 8 kW power at...

Key Concepts

AC Circuit Analysis Using Phasors

AC circuit analysis using phasors involves representing sinusoidal voltages and currents as complex numbers, which simplifies the calculation of circuit parameters such as magnitude and phase relationships. This method is particularly useful in multi-phase power systems, as it allows for the analysis of balance, load impedance, and the impact of reactive components on overall system performance. Phasor analysis bridges the gap between time-domain behavior and the algebraic manipulation of circuit equations, making it a cornerstone technique in electrical engineering.

Power Factor

Power factor is a measure of how effectively electrical power is being converted into useful work output, defined as the cosine of the phase angle between the voltage and current waveforms. In AC circuits, it indicates the proportion of power that is used to perform work, with the remainder being stored temporarily in the system’s reactive components. A lower power factor implies that more current is required to deliver the same amount of useful power, resulting in increased losses in the system.

Inductive Impedance

Inductive impedance represents the opposition that an inductor offers to the flow of alternating current, due to its inherent property of inductance. It is defined by the product of the angular frequency and the inductance, expressed as a reactance, and is a central concept when analyzing circuits that contain coils. Inductive loads are characterized by a phase shift where the current lags behind the voltage, which is critical in the calculation of load parameters in AC circuits.

Three-Phase AC Power

Three-phase AC power systems are used to deliver electrical energy efficiently by using three alternating currents that are offset in phase by 120 degrees. This system configuration allows for a constant power transfer, reduces conductor material, and improves the efficiency of power generation and distribution. It involves concepts such as real, reactive, and apparent power, which are essential for analyzing how loads behave in balanced three-phase networks.

Star (Wye) Connection

A star (or wye) connection is a method of connecting the load or source such that one end of each of the three components is connected to a common neutral point. In a star configuration, the phase voltage (voltage across each component) is lower than the line voltage (voltage between any two of the three lines), and the relationship between them is determined by a constant factor. This connection is widely used in three-phase systems owing to its efficiency and the ease of obtaining dual voltage levels for different applications.

Transcript

00:01 Here given that we have a star connected inductor whose draw 8 kilowatt power at 0 .8 power factor from the supply 3 phase 415 volt and 50 hertz.

00:13 And in this question we will find the line currents that is ia, ib and ic and the parameter of inductor.

00:22 That means the resistance of this inductor and the inductance of this inductor.

00:28 So let's start.

00:31 We assume that z is the impedance of this inductor.

00:36 And here i represent the system by using given data.

00:42 Here we have a three -phase 50 hertz, 415 volt ac supply.

00:48 And here this is a star -connected inductor.

00:53 Whose per -phase impedance is z.

00:57 Here abc shows phase a, phase a, phase b and phase c.

01:01 Here van is the phase voltage and n is the neutral point.

01:06 Now in step first we find van that is phase voltage and we know phase voltage is equal to line voltage divide by root 3.

01:14 Here i take van as reference so van become equal to 415 divided by root 3 angle 0 degree volt.

01:25 Here active power drawn by the inductor is equal to 8 kilowatt and we know power drawn by by a 3 phase system is given by p equals to root 3 vl i .l cos phi here p is equal to 8 k means 8 into 10 to the power 3 is equal to root 3 vl is equal to 415 is unknown and the value of cos phi is equal to 0 .8 now by solving this we obtain the value of i .l is equal to 13 point nine one two amper here this is the magnitude of il so il in vector form will become equals to 13 .912 angle negative cos inverse 0 .8 here negative sign comes because il lakes v an angle cos inverse 0 .8 here the value of i is equal to ia and this will become equal to 13 .912 angle negative 38 .86 degree ampere...

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