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31\. 72 . An \(L-R-C\) series circuit 220 \(\mathrm{W}\) from a \(120-\mathrm{V}(\mathrm{rms})\) 50.0-Hz ac line. The power factor is 0.560 , and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacitor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

Short Answer

Expert verified
(a) Net resistance is determined by dividing real power by square of current. (b) Correcting for unity power factor requires capacitive reactance to match inductive reactance. (c) Power drawn is same as apparent power with unity power factor.

Step by step solution

01

Calculate the Apparent Power

Given the real power \( P = 220 \text{ W} \) and the root mean square (rms) voltage \( V = 120 \text{ V} \), the apparent power \( S \) can be calculated by dividing the real power by the power factor \( \text{pf} = 0.560 \). Thus, \( S = \frac{P}{\text{pf}} = \frac{220}{0.560} \).
02

Find the Total Current

Using the apparent power calculated in Step 1 and the voltage, calculate the current \( I \) as follows: \( I = \frac{S}{V} \).
03

Determine the Resistance

Using the power formula \( P = I^2 R \) and rearranging for \( R \), we find \( R = \frac{P}{I^2} \).
04

Calculate Inductive Reactance and Impedance

Given that the source voltage leads the current, the circuit has inductive reactance. Calculate the impedance \( Z \) using \( Z = \frac{V}{I} \), and then the inductive reactance \( X_L = \sqrt{Z^2 - R^2} \).
05

Find Capacitance for Power Factor Unity

To achieve a power factor of unity, the inductive and capacitive reactances must cancel each other out, \( X_C = X_L \). Hence, the capacitance \( C \) can be found using \( X_C = \frac{1}{2\pi f C} \) and rearranging for \( C \): \( C = \frac{1}{2\pi f X_L} \).
06

Calculate New Power Drawn

With a power factor of 1 and new capacitance in place, the power \( P' \) equals the apparent power of the corrected circuit. Therefore, \( P' = V \cdot I \).

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Key Concepts

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

Power Factor
The power factor in an electrical circuit is a measure that indicates how effectively the circuit is converting electrical power into useful work output. It is defined as the ratio of real power, which performs actual work, to apparent power, which is the product of the current and voltage supplied to the circuit. A power factor of 1 (or unity) means that all the power is being efficiently used, with minimum loss. When the power factor is less than 1, as in this exercise (0.560), this indicates inefficiencies and a portion of the energy is not contributing to useful work but instead may be stored, dissipated, or reflected back in the circuit.
  • A power factor is a dimensionless number between 0 and 1.
  • It can be calculated as the cosine of the phase angle between the current and voltage.
  • Improving a low power factor increases the efficiency of the power system.
Apparent Power
Apparent power, denoted as \(S\), represents the total power that flows to a device or system. Unlike real power, apparent power does not necessarily do work; it's a combination of real power and reactive power. In an L-R-C series circuit, apparent power is calculated by dividing the real power by the power factor.
If we apply this to our example:
  • Real power \(P = 220 ext{ W}\)
  • Power factor \(\text{pf} = 0.560\)
Using these values, the apparent power can be calculated:\[S = \frac{P}{\text{pf}} = \frac{220}{0.560}\]The result gives an insight into how much power is "apparently" consumed under given conditions, though it might not be fully utilized for performing work.
Capacitive Reactance
Capacitive reactance is the resistance that a capacitor offers to the change of voltage across it. It is a type of resistance that only appears in AC circuits because of the alternating nature of the voltage and current. This concept is vital for circuits where balancing inductive reactance is necessary, as seen when aiming for a power factor of unity.
The formula for capacitive reactance \(X_C\) is given by:\[X_C = \frac{1}{2\pi f C}\]where \(f\) is the frequency, and \(C\) is the capacitance.
In our exercise, capacitive reactance is used to balance the inductive reactance \(X_L\), which allows the circuit to operate at a power factor of unity. This balance is crucial to minimize losses and enhance efficiency. When \(X_C = X_L\), the reactive powers cancel each other out, leading to a purely resistive circuit state.
Inductive Reactance
Inductive reactance occurs when current flows through an inductor in an AC circuit. It creates resistance by inducing a back voltage as the current alternates, effectively opposing changes in current. This concept is critical in circuits where the source voltage leads the current, as it points towards the presence of inductive reactance.
The formula for finding inductive reactance \(X_L\) is in relation to the impedance \(Z\) and resistance \(R\) using:\[X_L = \sqrt{Z^2 - R^2}\]where \(Z \) is calculated by \(Z = \frac{V}{I} \) with \(V\) being the voltage and \(I\) the current.
In our exercise, the goal is to find the value of inductive reactance so we can effectively adjust it by adding a capacitor, achieving a power factor of unity. By nullifying the inductive reactance with capacitive reactance (\

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Most popular questions from this chapter

31.35. A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) capacitor, and a \(4.50-\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency; (b) twice the resonance angular frequency; (c) half the resonance angular frequency.

31.25. The power of a certain CD player operating at 120 \(\mathrm{V}\) rms is 20.0 \(\mathrm{W}\) . Assuming that the CD player behaves like a pure resistance, find (a) the maximum instantancous power; (b) the rms current; (c) the resistance of this player.

31.33. In an \(L . R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=0.0360 \mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

31.13. A \(150-\Omega\) resistor is connected in series with a \(0.250-\mathrm{H}\) inductor. The voltage across the resistor is \(v_{R}=(3.80 \mathrm{V}) \cos [(720 \mathrm{rad} / \mathrm{s}) t]\) (a) Derive an expression for the circuit current. (b) Determine the inductive reactance of the inductor. (c) Derive an expression for the voltage \(v_{L}\) across the inductor.

31.51. An \(L-R-C\) series circuit is connected to an ac source nf constant voltage amplitude \(V\) and variable angular frequency \(\omega\) . (a) Show that the current amplitude, as a function of \(\omega,\) is $$ I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}} $$ (b) Show that the average power dissipated in the resistor is $$ P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}} $$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C}\) ; that is, when the source frequency equals the resonance frequency of the circuit. (d) Graph \(P\) as a function of \(\omega\) for \(V=100 \mathrm{V}\) .

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