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In an \(L-R-C\) series circuit, the components have the following values: \(L = 20.0\space mH\), \(C = 140\space nF\), and R = 350 \(\Omega\). The generator has an rms voltage of 120 V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

Short Answer

Expert verified
(a) 10.59 W, (b) 10.59 W.

Step by step solution

01

Calculate Inductive Reactance

The inductive reactance \( X_L \) can be calculated using the formula \( X_L = 2\pi fL \). Given that \( f = 1.25 \) kHz and \( L = 20.0 \) mH, we find: \[ X_L = 2\pi \times 1250 \times 20 \times 10^{-3} = 157.08 \space \Omega \]
02

Calculate Capacitive Reactance

The capacitive reactance \( X_C \) is given by \( X_C = \frac{1}{2\pi fC} \). With \( f = 1.25 \) kHz and \( C = 140 \) nF, we have: \[ X_C = \frac{1}{2\pi \times 1250 \times 140 \times 10^{-9}} = 908.61 \space \Omega \]
03

Find Total Impedance

The total impedance \( Z \) in an L-R-C circuit is given by the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Substituting in the known values, \( R = 350 \space \Omega \), \( X_L = 157.08 \space \Omega \), \( X_C = 908.61 \space \Omega \):\[ Z = \sqrt{350^2 + (157.08 - 908.61)^2} = 689.95 \space \Omega \]
04

Determine Current in the Circuit

The current \( I \) is determined using the formula: \[ I = \frac{V_{rms}}{Z} \]Where \( V_{rms} = 120 \) V and \( Z = 689.95 \space \Omega \):\[ I = \frac{120}{689.95} = 0.174 \space \text{A} \]
05

Calculate Power Supplied by the Generator

The power supplied \( P \) can be found with the formula: \[ P = V_{rms} \cdot I \cdot \cos \phi \]Where \( \cos \phi = \frac{R}{Z} \) and \( R = 350 \space \Omega \), \( Z = 689.95 \space \Omega \):\[ \cos \phi = \frac{350}{689.95} = 0.507 \]Thus, \[ P = 120 \times 0.174 \times 0.507 = 10.59 \space \text{W} \]
06

Calculate Power Dissipated in the Resistor

The power dissipated in the resistor \( P_R \) is given by \[ P_R = I^2 R = (0.174)^2 \times 350 = 10.59 \space \text{W} \]

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Key Concepts

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

Inductive Reactance
Inductive reactance is a crucial concept when working with alternating current (AC) circuits involving inductors. It represents the opposition that an inductor offers to the change in current flow. The formula to calculate inductive reactance is given by:\[ X_L = 2\pi fL \]where:
  • \( X_L \) is the inductive reactance measured in ohms (\( \Omega \))
  • \( f \) is the frequency of the AC supply in hertz (Hz)
  • \( L \) is the inductance of the inductor in henrys (H)
In the exercise, you need to determine the inductive reactance by plugging in the values for frequency (1.25 kHz) and inductance (20 mH). The calculation shows that the inductor creates significant opposition, indicating how it influences the circuit's response to the AC voltage.
Capacitive Reactance
Capacitive reactance is equally important and describes the opposition to current flow posed by a capacitor in an AC circuit. This concept is essential for understanding how capacitors influence the phase and magnitude of circuit currents. The capacitive reactance can be calculated using:\[ X_C = \frac{1}{2\pi fC} \]where:
  • \( X_C \) is the capacitive reactance in ohms (\( \Omega \))
  • \( f \) is the frequency in hertz (Hz)
  • \( C \) is the capacitance in farads (F)
In the example, substituting the frequency and capacitance values yields a capacitive reactance that is much larger than the inductive reactance. This disparity is essential when analyzing the total impedance in the circuit. Similarly, understanding the lead-lag relationship between current and voltage across a capacitor helps in visualizing the behavior in AC applications.
Impedance Calculation
Impedance is a vital parameter in L-R-C circuits as it combines the effects of resistance, inductive reactance, and capacitive reactance. It provides a comprehensive measure of the circuit's total opposition to AC electricity. The impedance is often calculated using the formula:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]where:
  • \( Z \) is the total impedance (\( \Omega \))
  • \( R \) is the resistance (\( \Omega \))
  • \( X_L \) and \( X_C \) are the inductive and capacitive reactances respectively
For this calculation, the relative magnitudes of \( X_L \) and \( X_C \) greatly affect the result of \( Z \). A larger capacitive reactance than inductive reactance suggests a net capacitive behavior, impacting the circuit's phase angle and power factor.
Power Dissipation in Resistors
Power dissipation in resistors is a fundamental concept, especially in AC circuits like L-R-C circuits. Unlike reactance, power in resistors is dissipated as heat, indicating energy loss in the circuit. The power dissipated can be calculated by:\[ P_R = I^2 R \]where:
  • \( P_R \) is the power dissipated in the resistor (watts)
  • \( I \) is the current through the circuit (amps)
  • \( R \) is the resistance (\( \Omega \))
In the L-R-C circuit exercise, once the circuit current is known, one simply applies the formula to find the power dissipated across the resistor. Notably, this power loss matches the power supplied by the generator in this lossless circuit, illustrating the key resistor's role in AC power analysis.

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Most popular questions from this chapter

A 250-\(\Omega\) resistor is connected in series with a 4.80-\(\mu\)F capacitor and an ac source. The voltage across the capacitor is v\(_C\) = (7.60 V)sin[120 rad/s)t]. (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage v\(_R\) across the resistor.

A resistance \(R\), capacitance \(C\), and inductance \(L\) are connected in series to a voltage source with amplitude \(V\) and variable angular frequency \(\omega\). If \(\omega\) = \(\omega$$_0\) , the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across the inductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of \(R\), \(C\), \(L\), and \(V\).

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?

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

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