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The power dissipated in a series RCL circuit is \(65.0 \mathrm{W}\), and the current is 0.530 A. The circuit is at resonance. Determine the voltage of the generator.

Short Answer

Expert verified
The voltage of the generator is approximately 122.63 V.

Step by step solution

01

Understand the Formula for Power

In an electrical circuit, the power dissipated can be calculated using the formula: \[ P = I^2 R \]where:- \( P \) is the power in watts (W),- \( I \) is the current in amperes (A),- \( R \) is the resistance in ohms (Ω). We also know that at resonance, the resistance \( R \) is the total impedance \( Z \) of the circuit, which is purely resistive.
02

Calculate the Resistance at Resonance

Since we are given that the circuit is at resonance and we know the electrical power and current, we can rearrange the formula to solve for resistance \( R = Z \):\[ R = \frac{P}{I^2} \]Substitute the given values:\[ R = \frac{65.0}{(0.530)^2} \]
03

Perform the Calculation

Calculate the resistance by performing the arithmetic:\[ R = \frac{65.0}{0.2809} \approx 231.38 \, \text{Ω} \]Thus, the resistance \( R \) at resonance is approximately 231.38 Ω.
04

Calculate the Voltage of the Generator

Now, use Ohm's Law, which relates voltage \( V \), current \( I \), and resistance \( R \):\[ V = I \times R \]Substitute the known values:\[ V = 0.530 \, \text{A} \times 231.38 \, \text{Ω} \]
05

Solve for Voltage

Calculate the voltage across the circuit:\[ V = 122.63 \, \text{V} \]Thus, the voltage of the generator is approximately 122.63 volts.

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Key Concepts

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

resonance
In an electrical circuit, resonance occurs when the reactive elements—the inductor and capacitor—cancel each other out, making the circuit behave as if it only contains resistance. This condition is specific to RLC (resistor, inductor, capacitor) circuits, where the inductive reactance is equal to the capacitive reactance. At resonance, the impedance of the circuit is minimal, and it is purely resistive.
What this means is that the entire impedance, denoted by \( Z \), is the resistance itself, denoted by \( R \). Consequently, the reactive components do not affect the circuit's overall current flow. This balance also implies that the voltage and the current are in phase, offering optimal conditions for power transfer.
  • Resonance frequency can be calculated if the values of inductance and capacitance are known, using the formula: \( f_r = \frac{1}{2\pi \sqrt{LC}} \), where \( f_r \) is the resonance frequency, \( L \) is the inductance, and \( C \) is the capacitance.
  • At resonance, the energy oscillates between the inductor and capacitor with minimal outside losses.
Ohm's Law
Ohm's Law is a fundamental principle for understanding electrical circuits, and it conveys the relationship between voltage, current, and resistance. The law is expressed with the formula: \( V = I \times R \), where \( V \) represents the voltage (in volts), \( I \) stands for the current (in amperes), and \( R \) is the resistance (in ohms).
This law plays a crucial role in the given problem because it allows calculation of the voltage across the generator in the series RLC circuit using the known values of current and resistance. Since it provides such a clear method of calculating one if the other two values are known, it's an indispensable tool for circuit analysis.
  • Ohm’s Law is applicable in a variety of circuit conditions, as long as the circuit components remain linear and the temperature is constant.
  • The law not only applies to poor conductors and resistors but also conductive solutions and gases.
impedance
Impedance is a broader concept than resistance as it takes into consideration not only the resistive attributes but also the reactive components such as capacitance and inductance. In alternating current (AC) circuits, impedance, denoted by \( Z \), is represented by the formula \( Z = R + jX \), where \( j \) is the imaginary unit and \( X \) stands for the reactance.
In a series RLC circuit, impedance plays a significant role since it determines the current flow and phase angle between current and voltage. However, at resonance, the impedance is purely resistive and is numerically equal to the resistance \( R \) of the circuit.
  • Impedance is measured in ohms (Ω) like resistance, but it involves both real and imaginary components.
  • In practical applications, impedance can affect how signals are transmitted and received.
  • For AC systems, the total impedance affects how the system responds and performs under different conditions.

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Most popular questions from this chapter

A series \(\mathrm{RCL}\) circuit contains a \(47.0-\Omega\) resistor, a \(2.00-\mu \mathrm{F}\) capacitor, and a \(4.00-\mathrm{mH}\) inductor. When the frequency is \(2550 \mathrm{Hz},\) what is the power factor of the circuit?

Two parallel plate capacitors are filled with the same dielectric material and have the same plate area. However, the plate separation of capacitor 1 is twice that of capacitor 2. When capacitor 1 is connected across the terminals of an ac generator, the generator delivers an rms current of 0.60 A. Concepts: (i) Which of the two capacitors has the greater capacitance? (ii) Is the equivalent capacitance of the parallel combination \(\left(C_{\mathrm{P}}\right)\) greater or smaller than the capacitance of capacitor \(1 ?\) (iii) Is the capacitive reactance of \(C_{\mathrm{P}}\) greater or smaller than for \(C_{1} ?\) (iv) When both capacitors are connected in parallel across the terminals of the generator, is the current from the generator greater or smaller than when capacitor 1 is connected alone? Calculations: What is the current delivered by the generator when both capacitors are connected in parallel across the terminals?

A \(30.0-\mathrm{mH}\) inductor has a reactance of \(2.10 \mathrm{k} \Omega .\) (a) What is the frequency of the ac current that passes through the inductor? (b) What is the capacitance of a capacitor that has the same reactance at this frequency? The frequency is tripled, so that the reactances of the inductor and capacitor are no longer equal. What are the new reactances of (c) the inductor and (d) the capacitor?

Part \(a\) of the figure shows a heterodyne metal detector being used. As part \(b\) of the figure illustrates, this device utilizes two capacitor/ inductor oscillator circuits, A and B. Each produces its own resonant frequency, \(f_{0 \mathrm{A}}=1 /\left[2 \pi\left(L_{\mathrm{A}} C\right)^{1 / 2}\right]\) and \(f_{0 \mathrm{B}}=1 /\left[2 \pi\left(L_{\mathrm{B}} C\right)^{1 / 2}\right]\). Any difference between these frequencies is detected through earphones as a beat frequency \(\mid f_{0 \mathrm{B}}-\) \(f_{0 A} \mid .\) In the absence of any nearby metal object, the inductances \(L_{\mathrm{A}}\) and \(L_{\mathrm{B}}\) are identical. When inductor \(\mathrm{B}\) (the search coil) comes near a piece of metal, the inductance \(L_{\mathrm{B}}\) increases, the corresponding oscillator frequency \(f_{\mathrm{oB}}\) decreases, and a beat frequency is heard. Suppose that initially each inductor is adjusted so that \(L_{\mathrm{B}}=L_{\mathrm{A}},\) and each oscillator has a resonant frequency of \(855.5 \mathrm{kHz}\). Assuming that the inductance of search coil \(\mathrm{B}\) increases by \(1.000 \%\) due to a nearby piece of metal, determine the beat frequency heard through the earphones.

An ac series circuit has an impedance of \(192 \Omega,\) and the phase angle between the current and the voltage of the generator is \(\phi=-75^{\circ} .\) The circuit contains a resistor and either a capacitor or an inductor. Find the resistance \(R\) and the capacitive reactance \(X_{\mathrm{C}}\) or the inductive reactance \(X_{\mathrm{L}}\) whichever is appropriate.

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