Calculate the voltage 𝐕s that must be supplied to obtain 2 kW, 240 ∠0^∘ V rms, and a power fact

Calculate the voltage 𝐕s that must be supplied to obtain 2...

Key Concepts

Impedance Analysis

Impedance is the total opposition that a circuit presents to the flow of alternating current and includes both resistance (real component) and reactance (imaginary component). Understanding and calculating impedance is vital in determining how voltages and currents are distributed in a circuit, including how the supply voltage must be adjusted to meet specific load requirements.

Power Factor

The power factor is the ratio of real power to apparent power in an AC circuit and indicates how effectively electrical power is being converted into useful work. A power factor of 0.8 leading suggests that the circuit has a capacitive nature where the current leads the voltage, which is crucial for optimizing and correcting the load performance in an AC power system.

Complex Power

Complex power combines real power (the actual power consumed), reactive power (the power stored and released by inductors and capacitors), and apparent power (the product of rms voltage and current) into a single expression. It is essential for accurately characterizing the power behavior in AC circuits and ensuring that both energy consumption and storage effects are considered.

Phasor Representation

Phasors are used to represent sinusoidal functions in the frequency domain as complex numbers with a magnitude and an angle. This representation greatly simplifies calculations in AC circuits by converting differential equations into algebraic equations, making it easier to analyze voltage and current relationships.

AC Circuit Analysis

AC circuit analysis involves working with time-varying voltages and currents, typically represented by sinusoidal functions. By using techniques such as phasor representation and complex numbers, the analysis simplifies the treatment of circuits with resistors, capacitors, and inductors, enabling the use of algebraic methods to solve for voltages and currents.

Transcript

00:01 In this problem i can write the value of pa by pfa is equal to s a which is equal to 10 multiplication 10 to the power 3 by 0 .9 which is equal to 11 .1 to multiplication 10 to the power 3 which is equal to 11 .1 to k b a now the value of theta is equal to cos inverse pf a which is equal to to cos in verse 0 .9 which is equal to 25 .842 degree.

00:41 Going forward and writing the value of qa which is equal to s a sine theta a which is equal to 11 .12 kva multiplication sine 25 .842 degree which is equal to 4 .85 multiplication 10 to the power 3 .85 multiplication 10 to the power 3.

01:04 Which is equal to 4 .8 5 k v .a r now i can write the value of p b is equal to s b cos theta b which is equal to s b b f b which is equal to 15 multiplication 10 to the power 3 multiplication 0 which is equal to 12 multiplication 10 to the power 3 w which is equal to 12 k w now calculating the value of theta b which is equal to cos inverse pfb which is equal to cos inverse 0 .8 which is equal to 36 .87 degree the value of q b is equal to s b sine ceta b which is equal to 15 k b multiplication sine 36 .87 degree which is equal to 9 multiplication 10 to the power 3 so change the unit i can write it as 9 kv a r now the value of ps is equal to pa plus p b which is equal to 10 plus 12 which is equal to 22 kw the value of qs is equal to qa plus qb which is equal to 4 .85 plus 9 which is equal to 13 .85 kv a r now the value of ss apparent power delivered by source is equal to under root ps square plus qs square which is equal to under root 22 multiplication 10 to the power 3 whole square plus 13 .85 multiplication 10 to the power 3 square which is equal to 26 kva as the value.

03:16 Now drawing the power triangle for the source.

03:20 So just look at it carefully...

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