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A series RCL circuit is at resonance and contains a variable resistor that is set to \(175 \Omega\). The power delivered to the circuit is \(2.6 \mathrm{~W}\). Assuming that the voltage remains constant, how much power is delivered when the variable resistor is set to \(5620 ?\)

Short Answer

Expert verified
The new power delivered is approximately 0.081 W.

Step by step solution

01

Determine the relationship between power and resistance at resonance

In a series RCL circuit at resonance, the power delivered to the circuit is given by the formula: \[ P = \frac{V^2}{R} \]Where \( P \) is the power, \( V \) is the voltage, and \( R \) is the resistance. At resonance, this implies that the power is inversely proportional to the resistance.
02

Calculate the voltage across the circuit using initial conditions

Using the power formula, with initial conditions set at \(175 \Omega\) and power \(2.6 \mathrm{~W}\), we can determine the voltage:\[ P = \frac{V^2}{175} = 2.6 \]Rearrange for \( V^2 \):\[ V^2 = 2.6 \times 175 \]Then calculate \( V^2 \): \[ V^2 = 455 \] Thus, \( V = \sqrt{455} \).
03

Use V to calculate the new power when R is 5620

Now that we know \( V^2 = 455 \), use this to find the power delivered when \( R = 5620 \Omega \): \[ P_{new} = \frac{V^2}{5620} \]Substitute for \( V^2 \):\[ P_{new} = \frac{455}{5620} \]Calculate \( P_{new} \):\[ P_{new} = 0.081 \mathrm{~W} \].

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Key Concepts

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

Power Calculation
In a series RLC circuit at resonance, the power delivered to the circuit leverages a fundamental concept of electrical engineering: the relationship between power, voltage, and resistance. This relationship is defined by the formula:
  • \( P = \frac{V^2}{R} \)
where:- \( P \) is the power in watts,- \( V \) is the voltage in volts, and- \( R \) is the resistance in ohms.
This equation shows an inverse relationship between power and resistance, meaning that as resistance increases, power decreases, assuming a constant voltage.
Understanding this principle is crucial for calculating power in varying resistance settings, particularly in circuits designed to operate at resonance—where impedance is minimized, and power transfer efficiency is maximized.
Resistance Effect on Power
The effect of resistance on power is a pivotal concept when analyzing circuits at resonance. At resonance, the circuit assumes a simplified form, as the impedance contributed by inductance and capacitance cancels out, highlighting the importance of resistance.
In a series RLC circuit, by increasing the resistance, the power delivered decreases. Conversely, less resistance would result in higher power delivery. This is due to the formula \( P = \frac{V^2}{R} \) which directly ties resistance to its effect on power.
  • For higher resistance levels (e.g., 5620 \( \Omega \)), power delivery drops as more voltage is needed to overcome that resistance.
  • In lower resistance scenarios (e.g., 175 \( \Omega \)), more power is delivered for the same voltage.
Understanding this relationship allows engineers to design circuits with desired power outputs by wisely adjusting resistance levels.
Series RLC Circuit
A series RLC circuit comprises a resistor (R), an inductor (L), and a capacitor (C) connected in series. These components each contribute to the circuit's total impedance, which affects how power is distributed.
At resonance, the inductive and capacitive reactances cancel each other out, resulting in a purely resistive circuit. This unique condition maximizes the voltage and current being in phase, allowing for optimal power transfer.
The resonance condition can be exploited in applications such as tuning filters and radios where a specific frequency is needed. The inherent simplicity that occurs at resonance makes calculations straightforward by focusing primarily on resistive factors, gasoline the understanding of how circuits behave at different frequencies.
Ohm's Law in Circuits
Ohm's Law is a core principle in electrical engineering and is expressed mathematically as:
  • \( V = IR \)
where:- \( V \) stands for voltage in volts,- \( I \) represents current in amperes, and- \( R \) denotes resistance in ohms.
This fundamental relationship allows us to predict how changes in resistance or current can affect the circuit's behavior and performance, especially at resonance.
In our context of a series RLC circuit at resonance, maintaining a constant voltage and varying the resistance directly alters the current. This alteration then influences the power output as described in previous concepts. Ohm's Law serves as a guiding principle for diagnosing and understanding electrical circuits, making it indispensable for any circuit analysis.

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Most popular questions from this chapter

A capacitor (capacitance \(C_{1}\) ) is connected across the terminals of an ac generator. Without changing the voltage or frequency of the generator, a second capacitor (capacitance \(C_{2}\) ) is added in series with the first one. As a result, the current delivered by the generator decreases by a factor of three. Suppose the second capacitor had been added in parallel with the first one, instead of in series. By what factor would the current delivered by the generator have increased?

When a resistor is connected across the terminals of an ac generator \((112 \mathrm{~V})\) that has a fixed frequency, there is a current of \(0.500 \mathrm{~A}\) in the resistor. When an inductor is connected across the terminals of this same generator, there is a current of \(0.400 \mathrm{~A}\) in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what is (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?

Multiple-Concept Example 3 reviews some of the concepts needed for this problem. An ac generator has a frequency of \(4.80 \mathrm{kHz}\) and produces a current of \(0.0400 \mathrm{~A}\) in a series circuit that contains only a \(232-\Omega\) resistor and a \(0.250-\mu F\) capacitor. Obtain (a) the voltage of the generator and (b) the phase angle between the current and the voltage across the resistor/capacitor combination.

Concept Questions Part \(a\) of the drawing shows a resistor and a charged capacitor wired in series. When the switch is closed, the capacitor discharges as charge moves from one plate to the other. Part \(b\) shows a plot of the amount of charge remaining on each plate of the capacitor as a function of time. (a) What does the time constant \(\tau\) of this resistor-capacitor circuit physically represent? (b) How is the time constant related to the resistance \(R\) and the capacitance \(C ?\) (c) In part \(c\) of the drawing, the switch has been removed and an ac generator has been inserted into the circuit. What is the impedance \(Z\) of this circuit? Express your answer in terms of the resistance \(R,\) the time constant \(\tau,\) and the frequency \(f\) of the generator. Problem The circuit elements in the drawing have the following values: \(R=18 \Omega\) \(V_{\mathrm{rms}}=24 \mathrm{~V}\) for the generator, and \(f=380 \mathrm{~Hz}\). The time constant for the circuit is \(\tau=3.0 \times 10^{-4} \mathrm{~s} .\) What is the rms current in the circuit?

A circuit consists of a resistor in series with an inductor and an ac generator that supplies a voltage of \(115 \mathrm{~V}\). The inductive reactance is \(52.0 \Omega,\) and the current in the circuit is 1.75 A. Find the average power delivered to the circuit.

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