(a) Compute the impedance of a series R-L-C circuit at angular frequencies of 1000,750, and 500

(a) Compute the impedance of a series R-L-C circuit at...

Key Concepts

Phasor Diagram

A phasor diagram is a graphical tool used to represent the magnitude and phase relationships of sinusoidal functions in AC circuits. By plotting voltage and current as vectors (phasors) in the complex plane, the diagram aids in visualizing how resistance and reactance combine to produce the overall impedance, as well as how the current leads or lags the voltage based on the phase angle.

Phase Angle

The phase angle in an AC circuit represents the difference in phase between the applied voltage and resulting current. It is determined by the ratio of the net reactive component to the resistive component of the impedance. Understanding the phase angle is essential for analyzing the time relationship between voltage and current, which has practical implications in power delivery and energy efficiency.

Angular Frequency

Angular frequency, denoted by ? and measured in radians per second, is a measure of how quickly an alternating signal oscillates. In the context of AC circuits, it directly influences the reactance of inductors and capacitors, thereby affecting the impedance and the overall dynamic behavior of the circuit, including resonance and phase shifts.

Impedance

Impedance is a fundamental concept in alternating current (AC) circuit analysis that quantifies the total opposition a circuit presents to the flow of current when driven by a sinusoidal source. It is a complex quantity combining resistance (the real part) and reactance (the imaginary part) and varies with frequency, making it essential for understanding how circuits behave under varying AC conditions.

Series RLC Circuit

A series RLC circuit consists of a resistor, an inductor, and a capacitor connected in series. This configuration is significant because it encapsulates the interplay between resistive and reactive elements in AC circuits. The dynamic interplay of inductive and capacitive reactance with resistance gives rise to phenomena such as resonance, frequency-dependent impedance, and phase differences between current and voltage.

Reactance

Reactance is the component of impedance that arises due to the presence of inductors and capacitors in an AC circuit. It is frequency-dependent, with inductive reactance increasing with frequency and capacitive reactance decreasing with frequency. The net reactance, the difference between inductive and capacitive reactances, plays a critical role in determining both the magnitude and phase angle of the overall impedance in AC circuits.

Transcript

00:02 Hi, in the first part of this given problem in an rlc series circuit, the value of resistance is 200 ome self -inductance is 0 .90 henry and the capacitance is having a value of 2 .00 micro -farrid.

00:24 Now first of all the frequency, the angular frequency is given as 1 ,000 gradient per second at this much angular frequency we have to find impedance of the circuit for which first of all we find the inductive reactance which is xl is equal to omega into l means this is 1 ,000 multiplied by 0 .9 which comes out to be 900 oom then we find capacity reactance xc which is equal to 1 by omega c means this is 1 by 1000 into c means 2 micro ferret or 2 into 10 dash power minus 6 and here it comes out to be 500 oom so the impedance at this much frequency omega is equal to 1 ,000 radian per second will be given by square root of x xl minus xc the whole square plus r square means net opposition offered by reactants and resistance.

01:34 So this is 900 minus 500 to the whole square plus 200 to the whole square.

01:44 This is the square of 400 square root of square of 400 plus square of 200, which will come out to be 16000000 plus 4 .000.

01:59 0 0.

02:01 So it comes out to be 20 0000 which is finally z is equal to 447 .2 ome.

02:13 First answer for the first part of this problem.

02:18 Now in the same first part of the problem this time the angular frequency has been reduced to 750 radian per second so now using the same expressions first of all we find inductive reactance xl is equal to omega l means this is 750 multiplied by 0 .9 and it comes out to be 700, 675 om and then capacity reactants xc is equal to 1 by omega c is equal to 1 by 750 into 2 into 10 dash bar minus 6 and here it comes out to be 666 .7 home.

03:03 So finally impedance now in this case will be x xl minus xc again means 675 minus 666 .7 to the whole square plus resistance which is 200 to the whole square.

03:21 So, here it will come out to be 8 .3 to the whole square plus 200 to the whole square means this is 4 .0069 .44 which comes out to be 200 .17 ome.

03:41 This is second answer for the first part of the problem.

03:46 Impedance of the circuit when the frequency is decreased to 750 gradient per second the impedance has also been reduced now in the same first part of the problem now the angular frequency is further reduced to just 500 radiant per second to this time xl is equal to omega l again 500 into 0 .9 means this is 450 oom and the capacitive reactants 1 by omega c means 1 by 500 into 2 into 10 dash per minus 6 and it comes out to be 1 ,000 ome means this time the passive reactants is more than the inductive so the impedance this time will be xc minus xl to the whole square plus r square means this is 1000 minus 450 to the whole square plus 200 the whole square means this is 302 500 plus 400 the whole square means this is 302 500 plus 4 0 0 0 0 0 0 so it comes out to be 3 4 4 to 500 and finally this time impedance is 585 .2 home means the impedance again starts increasing so in this case we have seen in all these three parts first of all the impedance was reducing and then it increases again if we keep on decreasing the angular frequency now in the second part of the problem we have to find that trend of change in the amplitude of the current.

05:49 So as amplitude of the current is given as the amplitude of emf applied divided by the impedance and as impedance first decreases and then increases again increases again and current is depending inversely.

06:17 On impedance so the amplitude of current amplitude of current will first increase and then will decrease now in the third part of the problem we have to find the phase difference between the current and voltage at the angular frequency 1000 radiant per second so that will be given by 10 5 is equal to xl minus xc the net reactants divided by resistance and at a thousand radian per second angular frequency xl was 900 om xc was 500 om and resistance is having the fixed value of 200 om it comes out to be 400 by 200 means this is 2 the phase difference is is 10 inverse 2, which comes out to be 63 .4 degree approximately.

07:36 Answer for the third part of the problem...

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