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Two \(660 \mathrm{V}\) (rms) loads are connected in parallel. The two loads draw a total average power of \(52,800 \mathrm{W}\) a power factor of 0.8 leading. One of the loads draws \(40 \mathrm{kVA}\) at a power factor of 0.96 lagging. What is the power factor of the other load?

Short Answer

Expert verified
The power factor of the other load is 0.5538 leading.

Step by step solution

01

Calculate Total Apparent Power

First, we calculate the total apparent power drawn by the loads. Given the total average power is \(52,800 \mathrm{W}\) and the power factor is 0.8 (leading), we can use the formula: \[ S_{\text{total}} = \frac{P_{\text{total}}}{\text{Power Factor}} = \frac{52,800}{0.8} = 66,000 \mathrm{VA}. \]
02

Identify Apparent Power for Load 1

One of the loads consumes \(40 \mathrm{kVA}\) at a power factor of 0.96 lagging. This means \(S_1 = 40,000 \mathrm{VA}\), and the active power (\(P_1\)) can be calculated using: \[ P_1 = S_1 \times \text{Power Factor} = 40,000 \times 0.96 = 38,400 \mathrm{W}. \]
03

Calculate Apparent and Active Power for Load 2

The apparent power of Load 2 \(S_2\) can be found by subtracting \(S_1\) from \(S_{\text{total}}\): \[ S_2 = S_{\text{total}} - S_1 = 66,000 - 40,000 = 26,000 \mathrm{VA}. \] The active power for Load 2 \(P_2\) is \(P_2 = P_{\text{total}} - P_1 = 52,800 - 38,400 = 14,400 \mathrm{W}.\)
04

Determine Power Factor of Load 2

The power factor of Load 2 (\( \text{Power Factor}_2 \)) can be determined by taking the ratio of the active power to the apparent power: \[ \text{Power Factor}_2 = \frac{P_2}{S_2} = \frac{14,400}{26,000} = 0.5538. \] Since the total power factor is leading while Load 1 has a lagging power factor, Load 2 must counteract with a leading power factor of approximately 0.5538.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Apparent Power
Apparent power is an important concept in electrical engineering, especially when dealing with AC circuits. It combines both active and reactive power, representing the total power flow in a circuit. Apparent power is measured in Volt-Amperes (\mathrm{VA}). In simple terms, it helps identify the capacity an electrical system can handle without considering the phase difference between voltage and current.
  • Apparent power (S) is calculated using the voltage (V) and the current (I): \( S = V imes I \) .
  • For systems with a given power factor, apparent power can also be determined using the formula \( S = \frac{P}{\text{Power Factor}} \), where P is the active power.
Understanding apparent power is crucial as it determines how much load a system can support. In the given exercise, the total apparent power was calculated to ensure the loads are within the system's capabilities.
Power Factor
Power factor is a measure of how effectively electrical power is being used. It's a ratio between active power, which does useful work, and apparent power. The power factor value ranges between -1 and 1 and it indicates the phase difference between voltage and current.
  • A power factor of 1 (or 100%) means all supplied energy is used effectively for work.
  • A low power factor suggests that more apparent power is required to perform the same amount of work.
  • If the current lags behind the voltage, the power factor is said to be lagging. If the current leads the voltage, the power factor is leading.
For the exercise, the power factor played a critical role in determining the unknown values. Knowing the power factor of one load and the total helped calculate the unknown power factor of the other load.
Active Power
Active power is the real power consumed in an electrical circuit. It's the portion of power that performs actual, useful work and is measured in Watts (\mathrm{W}). Active power does not get wasted on components like inductors and capacitors, which would otherwise draw power but not use it effectively for work.
  • The active power is directly used to perform the intended task in the circuit, like lighting or heating.
  • It can be calculated using the apparent power and the power factor as \( P = S \times \text{Power Factor} \).
Understanding active power assists in allocating adequate power for useful applications without surpassing designed limits. In the given scenario, calculating the active power for the second load, after knowing the first, played a vital role in determining the power factor for efficiency assessment.

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Most popular questions from this chapter

A group of small appliances on a \(60 \mathrm{Hz}\) system requires \(25 \mathrm{kVA}\) at 0.96 pf lagging when operated a \(125 \mathrm{V}(\mathrm{rms}) .\) The impedance of the feeder supplying the appliances is \(0.006+j 0.048 \Omega\). The voltage at the load end of the feeder is \(125 \mathrm{V}(\mathrm{rms})\). a) What is the rms magnitude of the voltage at the source end of the feeder? b) What is the average power loss in the feeder? c) What size capacitor (in microfarads) at the load end of the feeder is needed to improve the load power factor to unity? d) After the capacitor is installed, what is the rms magnitude of the voltage at the source end of the feeder if the load voltage is maintained at \(125 \mathrm{V}(\mathrm{rms}) ?\) e) What is the average power loss in the feeder for \((\mathrm{d}) ?\)

A load consisting of a 1350 \Omega resistor in parallel with a \(405 \mathrm{mH}\) inductor is connected across the ter minals of a sinusoidal voltage source \(v_{g}\), where \(v_{x}=90 \cos 2500 t \mathrm{V}\). a) What is the peak value of the instantaneous power delivered by the source? b) What is the peak value of the instantaneous power absorbed by the source? c) What is the average power delivered to the load? d) What is the reactive power delivered to the load? e) Does the load absorb or generate magnetizing vars? f) What is the power factor of the load? g) What is the reactive factor of the load?

Prove that if only the magnitude of the load impedance can be varied, most average power is transferred to the load when \(\left|Z_{\mathrm{L}}\right|=\left|Z_{\mathrm{Th}}\right| .\) (Hint: In deriving the expression for the average load power, write the load impedance \(\left(Z_{\mathrm{L}}\right)\) in the form \(Z_{\mathrm{L}}=\left|Z_{L}\right| \cos \theta+j\left|Z_{\mathrm{L}}\right| \sin \theta,\) and note that only \(\left|Z_{L}\right|\) is variable.

A dc voltage equal to \(V_{\mathrm{dc}} \mathrm{V}\) is applied to a resistor of \(R \Omega .\) A sinusoidal voltage equal to \(v_{s} V\) is also applied to a resistor of \(R \Omega\). Show that the de voltage will deliver the same amount of energy in \(T\) seconds (where \(T\) is the period of the sinusoidal voltage \()\) as the sinusoidal voltage provided \(V_{\mathrm{dc}}\) equals the rms value of \(v_{s}\). (Hint: Equate the two expressions for the energy delivered to the resistor.)

a) A personal computer with a monitor and keyboard requires \(60 \mathrm{W}\) at \(110 \mathrm{V}(\mathrm{rms})\). Calculate the rms value of the current carried by its power cord. b) A laser printer for the personal computer in (a) is rated at \(80 \mathrm{W}\) at \(110 \mathrm{V}(\mathrm{rms})\). If this printer is plugged into the same wall outlet as the computer, what is the rms value of the current drawn from the outlet?

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