/*! 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 57 A series \(R L C\) circuit consi... [FREE SOLUTION] | 91Ó°ÊÓ

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A series \(R L C\) circuit consists of an \(8.00-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a 50.0 \(\mathrm{-mH}\) inductor. A variable frequency source applies an emf of 400 V (rms) across the combination. Determine the power delivered to the circuit when the frequency is equal to half the resonance frequency.

Short Answer

Expert verified
The power delivered to the circuit at half the resonance frequency is given by \(P = I_{rms}^2 \times R\), after finding \(I_{rms}\) and using the resistance \(R = 8.00 \Omega\).

Step by step solution

01

Determine the Resonance Frequency

The resonance frequency, or resonant angular frequency, \(\omega_0\), of an RLC circuit is given by \(\omega_0 = \frac{1}{\sqrt{LC}}\) where L is the inductance and C is the capacitance. Calculate this frequency using the provided values, \(L = 50.0\times 10^{-3} H\) and \(C = 5.00\times 10^{-6} F\).
02

Calculate the Angular Frequency of the Source

Given that the frequency of the source is half the resonance frequency, the angular frequency of the source \(\omega\) is \(\omega = \frac{\omega_0}{2}\). Once \(\omega_0\) is found from Step 1, halve it to find \(\omega\).
03

Calculate the Impedance at Half the Resonance Frequency

The impedance \(Z\) at a given angular frequency \(\omega\) for a series RLC circuit is given by \(Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}\). Plug in the values of \(R\), \(L\), \(C\), and the \(\omega\) from Step 2 to calculate \(Z\).
04

Calculate the RMS Current in the Circuit

The RMS current \(I_{rms}\) can be found using the RMS voltage \(V_{rms}\) and the impedance \(Z\) with the formula \(I_{rms} = \frac{V_{rms}}{Z}\). Use the provided \(V_{rms} = 400V\) and the impedance found in Step 3 to calculate \(I_{rms}\).
05

Calculate the Power Delivered to the Circuit

The power delivered \(P\) to an RLC circuit is given by \(P = I_{rms}^2 \times R\), using the RMS current found in Step 4 and the resistance \(R\). This formula reflects that only the resistive part of the circuit dissipates energy as heat.

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Key Concepts

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

Resonance Frequency
When working with RLC circuits, resonance frequency is a pivot point. It's when the circuit is perfectly tuned, and the inductive and capacitive reactances counterbalance each other, leading to only the resistance dictating the impedance. This harmony happens at a specific frequency, denoted as the resonance frequency. To boil it down, it's calculated by the formula: \[\begin{equation} \(\omega_0 = \frac{1}{\sqrt{LC}}\) \end{equation}\] where L is the inductance in henrys (H) and C is the capacitance in farads (F).

In plain terms, imagine a swing—the resonance frequency is that perfect push that gets the swing to go higher with minimal effort. At resonance, the RLC circuit swings with the highest amplitudes of current, but with our exercise, the frequency we're dealing with is just half of this 'perfect push' point.
Impedance of RLC Circuit
Impedance in an RLC circuit is like the 'gatekeeper' of current. It's a combination of resistance and reactance and determines how much current is allowed through. Mathematically, it's characterized by the equation: \[\begin{equation} \(Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}\) \end{equation}\] Here, R is resistance, L is inductance, C is capacitance, and \(\omega\) is the angular frequency of the source. This might look complex, but really, it's just an expression of how much opposition there is to current flow at any given frequency.

Think of it as a party bouncer who decides who gets in (current) based on a specific rule (frequency). When the frequency is half the resonance, the bouncer (impedance) becomes more lenient, allowing a different amount of current to flow compared to at full resonance.
RMS Current
RMS current stands for Root Mean Square current. It's a way to express the effective current in AC circuits, like our RLC circuit. The RMS value equals the direct current that would deliver the same power to the resistor as the alternating current does over a complete cycle. To find the RMS current, we divide the RMS voltage by the impedance, following the equation: \[\begin{equation} \(I_{rms} = \frac{V_{rms}}{Z}\) \end{equation}\] RMS values are super useful because they give us a practical measure of how much current is actually doing the work in the circuit.

Let's use an everyday comparison: The RMS current is akin to the average amount of water flowing through a pipe, not the maximum capacity when it's full blast, which means it's a more realistic indication of continuous flow.
Power Delivered to RLC Circuit
Finally, let's talk about the power delivered to the RLC circuit. Power in this context is the amount of energy per unit time that's dissipated by the resistor as heat. The formula for power in terms of the other values we've talked about is: \[\begin{equation} P = I_{rms}^2 \times R \end{equation}\] Reflect on this: the power we measure here is the 'useful' work done, which, in electrical terms, equates to the heat dissipated by the resistor. The coils of a space heater glowing red hot is just like the resistor converting electrical power into heat in our RLC circuit.

In our exercise, we're taking the RMS current from earlier and squaring it, then multiplying it by the pure resistance to find how much power the circuit chews up. Note that this formula doesn’t consider power factor, as we assume the phase angle between voltage and current is zero, which is the case for a purely resistive load.

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Most popular questions from this chapter

An RLC circuit is used in a radio to tune into an FM station broadcasting at 99.7 \(\mathrm{MHz}\) . The resistance in the circuit is \(12.0 \Omega,\) and the inductance is 1.40\(\mu \mathrm{H}\) . What capacitance should be used?

An \(80.0-\Omega\) resistor and a \(200-\mathrm{mH}\) inductor are connected in parallel across a \(100-\mathrm{V}\) (rms), \(60.0-\mathrm{Hz}\) source. (a) What is the rms current in the resistor? (b) By what angle does the total current lead or lag behind the voltage?

An AC voltage of the form \(\Delta v=(100 \mathrm{V}) \sin (1000 t)\) is applied to a series \(R L C\) circuit. Assume the resistance is \(400 \Omega,\) the capacitance is \(5.00 \mu \mathrm{F},\) and the inductance is 0.500 \(\mathrm{H}\) . Find the average power delivered to the circuit.

Impedance matching. Example 28.2 showed that maximum power is transferred when the internal resistance of a DC source is equal to the resistance of the load. A transformer may be used to provide maximum power transfer between two AC circuits that have different impedances Z1 and Z2, where 1 and 2 are subscripts and the Z’s are italic (as in the centered equation). (a) Show that the ratio of turns \(N_{1} / N_{2}\) needed to meet this condition is $$\frac{N_{1}}{N_{2}}=\sqrt{\frac{Z_{1}}{Z_{2}}}$$ (b) Suppose you want to use a transformer as an impedance-matching device between an audio amplifier that has an output impedance of 8.00 \(\mathrm{k} \Omega\) and a speaker that has an input impedance of 8.00\(\Omega\) . What should your \(N_{1} / N_{2}\) ratio be?

A \(10.0-\Omega\) resistor, \(10.0-\mathrm{mH}\) inductor, and \(100-\mu \mathrm{F}\) capacitor are connected in series to a \(50.0-\mathrm{V}\) (rms) source having variable frequency. Find the energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency.

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