/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 An \(L-R-C\) series circuit draw... [FREE SOLUTION] | 91影视

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Chapter 31: Problem 63

An \(L-R-C\) series circuit draws 220 W from a 120-V (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) R = 32.89 惟; (b) C = 57.6 渭F; (c) P = 264 W.

Step by step solution

01

Understanding Power Factor

The power factor in an AC circuit is the cosine of the phase angle (\(\phi\)) between the source voltage and the current. It is given by \(\cos(\phi) = 0.560\). This implies \(\phi = \cos^{-1}(0.560)\).
02

Calculating Apparent Power

The apparent power \(S\) in an AC circuit is given by \(S = V_{rms} \times I_{rms}\), where \(V_{rms} = 120\, V\) and the real power \(P = 220\, W\). The relationship \(P = S \times \cos(\phi)\) can be rearranged to find \(S = \frac{P}{\cos(\phi)}\). Substitute \(P\) and \(\cos(\phi)\) to find \(S\).
03

Calculating RMS Current

Using \(S = V_{rms} \times I_{rms}\), where \(S\) is known from Step 2, solve for \(I_{rms}\): \(I_{rms} = \frac{S}{V_{rms}}\).
04

Finding Resistance R

The real power \(P\) can also be expressed as \(P = I_{rms}^2 \times R\). Rearrange this formula to find the resistance \(R = \frac{P}{I_{rms}^2}\). Substitute \(P\) and \(I_{rms}\) to find \(R\).
05

Calculating Capacitive Reactance for Unity Power Factor

To achieve a unity power factor, the capacitive reactance \(X_C\) must cancel out the inductive reactance \(X_L\). The condition for unity power factor is \(X_C = X_L - R \tan(\phi)\). Calculate \(X_L\) using \(X_L = \frac{V_{rms}}{I_{rms}} \sin(\phi)\), then find \(X_C\).
06

Determining Capacitance

Capacitive reactance \(X_C\) is given by \(X_C = \frac{1}{2\pi f C}\), where \(f = 50.0\, Hz\). Rearrange to solve for \(C\): \(C = \frac{1}{2\pi f X_C}\). Substitute values to find \(C\).
07

Calculating New Power Draw

With a power factor of unity, the new power draw \(P_{unity}\) is given by \(P_{unity} = V_{rms} \times I_{rms}\), where the new \(I_{rms}\) equals the original \(I_{rms}\) since the power factor is adjusted without changing supply parameters. Calculate \(P_{unity}\).

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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 is an important concept in an AC circuit. It is defined as the ratio of the real power to the apparent power. Mathematically, it's given as \[ \text{power factor} = \cos(\phi), \]where \( \phi \) is the phase angle between voltage and current. This phase angle is a result of the lagging or leading nature of the current with respect to the voltage, due to the presence of reactive components like inductors or capacitors. A power factor of 1 (unity) indicates all the power is consumed as real power, with no reactive power - this is the most efficient scenario for power use.
  • A low power factor implies inefficiency because more current is needed to deliver the same amount of real power.
  • In the exercise, getting from a power factor of 0.560 to unity means adjusting components to eliminate reactive power.
      • To improve power factor, you might need to adjust the circuit components, particularly capacitors, to balance out inductive effects.
Capacitive Reactance
Capacitive reactance is a measure of a capacitor's opposition to ac current. It is denoted by \( X_C \) and is inversely proportional to the frequency of the AC supply and the capacitance itself.The formula is:\[X_C = \frac{1}{2\pi f C},\]where \( f \) is the frequency and \( C \) is the capacitance.In an L-R-C series circuit:
  • The capacitive reactance helps to balance out the inductive reactance, \( X_L \), from inductors.
  • For the circuit to achieve a power factor of unity, \( X_C \) must exactly counteract the effects of \( X_L \).
Understanding how to calculate and apply capacitive reactance is key in tuning circuits for optimal performance. This is why the exercise focuses on adjusting capacitance to reach unity power factor.
Apparent Power
Apparent power \( S \) is the product of the root mean square (RMS) values of voltage and current in an AC circuit. It represents the total power flow but doesn't necessarily equate to usable power.The formula for apparent power is:\[S = V_{rms} \times I_{rms}.\]Key points include:
  • Apparent power consists of both real power and reactive power.
  • It's measured in volt-amperes (VA) and is always greater than or equal to real power.
  • In the exercise, apparent power helps determine currents and resistances leading to components' adjustments.
By understanding apparent power's role in circuits, students can better grasp why reactive components affect overall power consumption.
Real Power
Real power, often referred to as active power, is the component of power that is actually consumed or used in an AC circuit to perform work.The expression for real power \( P \) is:\[P = V_{rms} \times I_{rms} \times \cos(\phi).\]Important concepts about real power include:
  • It is measured in watts (W).
  • Represents the net work done by the circuit.
  • In the exercise, knowing the real power enables calculation of currents and resistance, setting groundwork for improving power factor through capacitive adjustments.
Real power distinguishes itself from apparent power by accounting only for efficiency - essentially, it's what power consumers want maximized for cost-effective operations.

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Most popular questions from this chapter

The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

A 200-\(\Omega\) resistor, 0.900-H inductor, and 6.00-\(\mu\)F capacitor are connected in series across a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad>s. (a) What are \(v, v_R, v_L\), and \(v_C\) at \(t = 20.0 ms\)? Compare \(v_R + v_L + v_C\) to \(v\) at this instant. (b) What are VR , VL, and VC? Compare V to \(V_R + V_L + V_C\). Explain why these two quantities are not equal.

An \(L-R-C\) series circuit has \(C\) = 4.80 \(\mu\)F, \(L\) = 0.520 H, and source voltage amplitude \(V\) = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of \(R\) for the resistor in the circuit?

(a) What is the reactance of a 3.00-H inductor at a frequency of 80.0 Hz? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 Hz? (c) What is the reactance of a 4.00-\(\mu\)F capacitor at a frequency of 80.0 Hz? (d) What is the capacitance of a capacitor whose reactance is 120 \(\Omega\) at 80.0 Hz?

A series ac circuit contains a 250-\(\Omega\) resistor, a 15-mH inductor, a 3.5-\(\mu\)F capacitor, and an ac power source of voltage amplitude 45 V operating at an angular frequency of 360 rad/s. (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?
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