/*! 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 33 For an \(R L C\) series circuit,... [FREE SOLUTION] | 91Ó°ÊÓ

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For an \(R L C\) series circuit, the voltage amplitude and frequency of the source are \(100 \mathrm{V}\) and \(500 \mathrm{Hz}\), respectively; \(R=500 \Omega ;\) and \(L=0.20 \mathrm{H} .\) Find the average power dissipated in the resistor for the following values for the capacitance: (a) \(C=2.0 \mu \mathrm{F}\) and (b) \(C=0.20 \mu \mathrm{F}.\)

Short Answer

Expert verified
The average power dissipated in the resistor for capacitance (a) \(C=2.0 \mu F\) is \(15.14 W\), and for capacitance (b) \(C=0.20 \mu F\) is \(2.88 W\).

Step by step solution

01

Identify the given values

In this problem, we have the following values: - Voltage amplitude: \(V_0 = 100 V\) - Frequency: \(f = 500 Hz\) - Resistance: \(R = 500 \Omega\) - Inductance: \(L = 0.20 H\) - Capacitance (a): \(C_1 = 2.0 \mu F\) - Capacitance (b): \(C_2 = 0.20 \mu F\)
02

Calculate the impedance for capacitance values

We know that the impedance for an RLC circuit is given by: \[Z = \sqrt{R^2 + (XL - XC)^2}\] Where \(XL = 2\pi f L\), is the inductive reactance, and \(XC = \frac{1}{2\pi f C}\), is the capacitive reactance. First calculate the inductive reactance and then for both capacitance values, calculate capacitive reactance and the impedance. For the inductive reactance: \[XL = 2\pi (500 Hz)(0.2 H) = 628.32 \Omega\] For capacitance \(C_1 = 2 \mu F\): \[XC_1 = \frac{1}{2\pi (500 Hz)(2 \cdot 10^{-6} F)} = 159.15 \Omega\] \[Z_1 = \sqrt{(500 \Omega)^2 + (628.32 \Omega - 159.15 \Omega)^2} = 573.60 \Omega\] For capacitance \(C_2 = 0.2 \mu F\): \[XC_2 = \frac{1}{2\pi (500 Hz)(0.2 \cdot 10^{-6} F)} = 1591.55 \Omega\] \[Z_2 = \sqrt{(500 \Omega)^2 + (628.32 \Omega - 1591.55 \Omega)^2} = 1318.64 \Omega\]
03

Calculate the current for both capacitance values

We can find the current using Ohm's law: \[I = \frac{V}{Z}\] For capacitance \(C_1 = 2 \mu F\): \[I_1 = \frac{100V}{573.60 \Omega} = 0.174 A\] For capacitance \(C_2 = 0.2 \mu F\): \[I_2 = \frac{100V}{1318.64 \Omega} = 0.076 A\]
04

Calculate the average power dissipated in the resistor

We can find the average power using the formula: \[P = I^2R\] For capacitance \(C_1 = 2 \mu F\): \[P_1 = (0.174 A)^2 (500 \Omega) = 15.14 W\] For capacitance \(C_2 = 0.2 \mu F\): \[P_2 = (0.076 A)^2 (500 \Omega) = 2.88 W\] So, the average power dissipated in the resistor for capacitance: (a) \(C=2.0 \mu F\) is \(15.14 W\), and (b) \(C=0.20 \mu F\) is \(2.88 W\).

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Key Concepts

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

Impedance Calculation in RLC Series Circuits
In an RLC series circuit, the concept of impedance is a central component that determines how the circuit reacts to an alternating current (AC). Impedance, denoted by the symbol \(Z\), is analogous to resistance in a direct current (DC) circuit, but it also takes into account the phase difference between the voltage and the current. The impedance is calculated using the formula:
\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]
where \(R\) is the resistance, \(X_L = 2\pi f L\) is the inductance reactance, and \(X_C = \frac{1}{2\pi f C}\) is the capacitive reactance. The terms \(X_L\) and \(X_C\) represent the opposition to the flow of current due to inductance and capacitance, respectively.

An important aspect to note is that inductance reactance increases with frequency, while capacitive reactance decreases with frequency. It is crucial to correctly calculate the magnitude of impedance, as it directly affects the AC current flowing in the circuit. In the given exercise, we find that changing the capacitance value has a significant impact on the impedance of the circuit.
Understanding Resonant Frequency
The resonant frequency of an RLC series circuit is the frequency at which the impedance is at its minimum, and consequently, the current is at its maximum. At the resonant frequency, the inductive reactance \(X_L\) and capacitive reactance \(X_C\) are equal, thus canceling each other out.
The formula to find the resonant frequency \(f_0\) is given by:
\[f_0 = \frac{1}{2\pi\sqrt{LC}}\]
At this frequency, the circuit behaves purely resistive, with the impedance equal to the resistance \(R\). This simplifies the analysis of the circuit since the phase angles between voltage and current are zero, making power calculations straightforward. While this exercise did not specifically ask for the calculation of the resonant frequency, understanding it is crucial for deeper insight into how RLC circuits behave and how they can be applied in practical scenarios like radio transmitters and tuners.
Average Power Dissipation in Resistors
Power dissipation in resistors is a measurement of how much electrical energy is converted to heat due to the resistance to current flow. In an RLC series circuit, the average power dissipated by a resistor can be determined using the RMS (root mean square) values of current or voltage. The formula to calculate the average power \(P\) in a resistor is:
\[P = I^2R\]
where \(I\) is the RMS current, and \(R\) is the resistance. This power denotes the energy used by the resistor every second and can be found by squaring the RMS current and then multiplying by the resistance.

Applying this to our exercise, we see that with lower impedance, the current through the resistor increases, leading to greater power dissipation. As the capacitance value changes from \(2.0 \mu F\) to \(0.20 \mu F\), the impedance increases, and the current decreases, resulting in lower power dissipation. The average power dissipation is then directly proportional to the square of the current flowing through the circuit for a given resistance.

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Most popular questions from this chapter

Neon signs require \(12-\mathrm{kV}\) for their operation. A transformer is to be used to change the voltage from \(220-\mathrm{V}\) (rms) ac to \(12-\mathrm{kV}\) (ms) ac. What must the ratio be of turns in the secondary winding to the turns in the primary winding? (b) What is the maximum rms current the neon lamps can draw if the fuse in the primary winding goes off at \(0.5 \mathrm{A} ?\) (c) How much power is used by the neon sign when it is drawing the maximum current allowed by the fuse in the primary winding?

A 1.5-k\Omega resistor and 30-mH inductor are connected in series, as shown below, across a \(120-\mathrm{V}\) (ms) ac power source oscillating at \(60-\mathrm{Hz}\) frequency. (a) Find the current in the circuit. (b) Find the voltage drops across the resistor and inductor. (c) Find the impedance of the circuit. (d) Find the power dissipated in the resistor. (e) Find the power dissipated in the inductor. (f) Find the power produced by the source.

The resonant frequency of an \(R L C\) series circuit is \(2.0 \times 10^{3} \mathrm{Hz} .\) If the self-inductance in the circuit is 5.0 \(\mathrm{mH},\) what is the capacitance in the circuit?

A resistor and inductor are connected in series across an ac generator. The emf of the generator is given by \(v(t)=V_{0} \cos \omega t, \quad\) where \(\quad V_{0}=120 \mathrm{V}\) and \(\omega=120 \pi \mathrm{rad} / \mathrm{s} ;\) also, \(R=400 \Omega\) and \(L=1.5 \mathrm{H}\) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the inductor.

What is the impedance of a series combination of a \(50-\Omega\) resistor, a \(5.0-\mu \mathrm{F}\) capacitor, and a \(10-\mu \mathrm{F}\) capacitor at a frequency of \(2.0 \mathrm{kHz}\) ?

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