A series circuit contains a 3.00 - H inductor, a 3.00 - mF capacitor, and a 30.0 - V resistor co

A series circuit contains a 3.00 - H inductor, a 3.00

Key Concepts

Impedance

Impedance in AC circuits is the total opposition that a circuit offers to alternating current, combining both resistance and reactance. In a series RLC circuit, impedance varies with frequency because the inductive reactance increases with frequency while the capacitive reactance decreases. Understanding impedance is crucial for determining how an AC circuit responds to different frequencies and for calculating parameters such as current and power.

AC Power Calculation

AC power calculation in circuits typically involves calculating the real power delivered, which is determined by the resistive element of the circuit. In series RLC circuits, real power is given by the formula P = V_rms^2 / R at resonance because the reactive components do not dissipate power. This concept is central to analyzing energy consumption in circuits and optimizing performance.

Series RLC Circuit

A series RLC circuit is an electrical circuit consisting of a resistor (R), inductor (L), and capacitor (C) connected in a single loop. The behavior of the circuit is influenced by the frequency of the applied AC source, which affects the circuit’s overall impedance due to the frequency-dependent reactances of the inductor and capacitor. This type of circuit is fundamental in AC analysis and finds applications in filtering, tuning and signal processing.

Resonance

Resonance in a series RLC circuit occurs when the inductive and capacitive reactances cancel each other out, resulting in the circuit having its minimum impedance, which is purely resistive. At this frequency, known as the resonance frequency, the voltage and current are in phase, and the circuit tends to draw maximum current. Resonance is a key concept in understanding and designing circuits for maximum power transfer and optimal performance.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that a series circuit contains an inductor whose inductance is equal to three henries.

00:12 The capacitance of the capacitor is equal to three micropharets.

00:22 And the resistor has a resistance equal to 30 oms.

00:33 And it is connected to an rms source of particle frequency that is equal to 120 volts.

00:45 So for par, we need to find the power deliberate to the circuit when the frequency of the source.

00:54 So for part a, when the frequency, it corresponds to the frequency, the resonance frequency.

01:06 Now, at resonance, we know that the reactance of the capacitance is equal to the reactance of the inductance.

01:19 So we will have, we know that the definition, of the reactance of the inductance is equal to omega -0, which is the angular frequency times the inductance.

01:34 And this is equal to the, in this case, we know that the angular frequency at resonance is going to be equal to 1 over the square root of the product between the inductance and the capacitance times the inductance.

01:52 So, simplifying this expression, we will find that this is the inductance divided by the capacitance, the square root of dots.

02:01 So we substitute those values in here.

02:05 So we will have that this is equal to three henries divided by the capacitance, which is three times 10 to the minus 6 ferrets.

02:17 So from this, we obtain 1 ,000 oms.

02:24 And so we are going to have that the impedance in this case is going to be equal to the resistance of the resistor.

02:38 So the maximum current in this case is going to be the product between, well, it's going to be the maximum the maximum or the rms value for the potential or the potential difference divided by the impudence.

03:04 So we will have that this is 120 balls divided by the impedance that in this case corresponds to the resistance of the resistor.

03:13 So that is 30 oms.

03:18 So from this we obtain a current of 4 umpers.

03:24 Then the power, the average power, the average power, is going to be equal to the maximum current to the square times the resistance.

03:39 So we substitute those two values in here we will have four ampers to the square times the resistance which is 30 oms.

03:49 So from this we obtain an average power of 480 bucks.

03:55 So that's a solution for part a of this problem.

04:00 Now for par p, we are told that now the frequency is one half the resonance frequency.

04:10 So we need to do a similar procedure as before, but in this case, the angular frequency is one half of the angular frequency for resonance.

04:22 So we're going to have that in this case, the, um, the, the induct, the, the, the, the, the, the reactance of the inductor is going to be half of the reactance of the inductor for omega zero.

04:44 So we're going to have that this is equal to simply 500 ums.

04:51 So we're also going to have in here that the reactance of the capacitor is going to be two times the reactance of the capacitor for omega -0, so we're going to have that this is 2 ,000 oms.

05:16 So in this case, the reactants, the impedance, sorry, is going to be equal to the square root of the resistance to the square, plus the difference between the reactance of the inductor minus the reactance of the capacitor, and that to the square.

05:37 And we take the square root of that.

05:38 So we need to substitute the values in here.

05:41 So we are going to have in here that this is the square root of 30 oms to the square plus 500 oms minus 2 ,000 oms and that to the square.

05:59 So from this, we take the square root of all this, and we obtained that the impedance of this circuit is 1 ,000, 1 ,500 oms.

06:13 Now, with this value, we can obtain the maximum current.

06:21 So it's going to be equal to 112 volts divided by the impedance, which is equal to 1 ,500 oms.

06:37 So from this, we obtain a maximum current equals to 0 .08 ampers.

06:44 So the average power in this case is going to be the current, the maximum current, times the 2nd square times the resistance.

06:55 So we will have in here 0 .08 umpers to the square times the resistance, which is 30 oms.

07:03 So from this we obtain an average power equals to 0 .192 watts.

07:10 So that's a solution for part b of this problem.

07:15 Now for par c, we are told that now the frequency of the source is one -fourth, the resonance frequency.

07:27 So we need to do something similar as before.

07:30 In this case, we will have that...

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