The circuit shown has a source voltage of 440 V rms resistance R=250 Ω, inductance L=0.800 H, an

step 1 In the given problem here this is a series RLC circuit. First of all, there is a resistance, the

The circuit shown has a source voltage of 440 V rms...

The circuit shown has a source voltage of $440 \mathrm{V} \mathrm{rms}$ resistance $R=250 \Omega,$ inductance $L=0.800 \mathrm{H},$ and capacitance $C=2.22 \mu \mathrm{F}$ (a) Find the angular frequency $\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: $V_{a b}, V_{b c}, V_{c d}, V_{b d},$ and $V_{a d}$ (d) The resistor is replaced with one of $R=125 \Omega .$ Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?

The circuit shown has a source voltage of $440 \mathrm{V} \mathrm{rms}$ resistance $R=250 \Omega,$ inductance $L=0.800 \mathrm{H},$ and capacitance $C=2.22 \mu \mathrm{F}$
(a) Find the angular frequency $\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: $V_{a b}, V_{b c}, V_{c d}, V_{b d},$ and $V_{a d}$ (d) The resistor is replaced with one of $R=125 \Omega .$ Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?

Key Concepts

Resonance in RLC Circuits

Resonance occurs in an RLC circuit when the reactive effects of the inductor and capacitor cancel each other out, meaning the inductive reactance ?L exactly equals the capacitive reactance 1?(?C). At this frequency the circuit’s impedance is minimized (in a series circuit) and the voltage across the components can exceed the source voltage even though the net reactive component is zero, a fact which is important when analyzing voltage distribution in the circuit.

Angular Frequency

Angular frequency, denoted by ? and measured in radians per second, is a key concept in analyzing AC circuits. It relates directly to the frequency at which the circuit operates and is used in computing reactive components, such as inductive reactance (?L) and capacitive reactance (1?(?C)). The resonance condition in an RLC circuit is expressed in terms of angular frequency through the formula ?0 = 1/?(LC).

Phasor Diagrams

Phasor diagrams are graphical representations used in AC circuit analysis to illustrate the magnitude and phase relationships between voltages and currents. In the context of a resonant RLC circuit, they help visualize that the resistor voltage is in phase with the current, while the inductor and capacitor voltages are 90 degrees out of phase relative to the current and cancel each other out at resonance.

Impedance in AC Circuits

Impedance is the total opposition that a circuit presents to alternating current, combining both resistance and reactance. In RLC circuits, impedance is calculated as a complex sum of the resistance (a real quantity) and the reactive impedances of the inductor and capacitor (imaginary quantities). At resonance in a series RLC circuit, the inductive and capacitive reactances cancel, leaving only the resistive component which simplifies analysis and current calculations.

Root Mean Square Values

Root Mean Square (RMS) values are used to express the effective magnitude of alternating voltages and currents. The RMS value of an AC quantity is equivalent to the DC value that would provide the same power dissipation in a resistor. This concept is essential when comparing or calculating voltages and currents in AC circuits since the instantaneous values vary sinusoidally.

Effects of Component Variations

Changing the components in an RLC circuit, such as the resistance, has significant effects on the circuit's behavior. While resistance affects the amplitude of the current and the damping of the circuit, the resonant frequency remains determined solely by the inductance and capacitance. Thus, replacing a resistor with a different value will alter the current level at resonance without changing the resonance frequency.

Transcript

00:01 In the given problem here this is a series rlc circuit.

00:07 First of all, there is a resistance, then there is an inductor coil, and then at last there is a capacitor all joint in series combination in the circuit, and then there this is the source of alternating voltage applied to it.

00:26 The terminals marked as a and b and c.

00:31 C and b here.

00:34 The value of rms potential applied is 440 volt.

00:40 The value of resistance is 250 ome.

00:44 Self -inductance is 0 .80 henry and capacitance is 2 .22 micro ferret.

00:57 Now in the first part of the problem we have to find resonating angular frequency which is given by the expression omega not is equal to 1 by and the root l into c so plugging in all known values here this omega not will be given by 1 divided by square root of self -inductance which is 0 0 0 0 multiplied by capacitance which is 2 .22 into 10 dash minus 6 which comes out to be 1 by 1 .33 into 10 dash par minus 3 radian per second and finally answer for the first part of the problem is approximately 750 radian per second which is the answer for the first part of the problem in the second part of the problem we have to draw a phaser diagram for the situation.

02:13 So to draw the phaser diagram, we show the rm's potential drop taking place across resistor along x -axis.

02:22 And that across inductor is shown along positive y -axis and that across the capacitor is shown along negative y -axis.

02:32 So here this is vr, this is vl, and this is vc.

02:38 First of all we find inductive reactance which will be equal to capacity reactants also and that will be given by the expression omega -note l for the inductive reactance this expression will be but actually it will be valid for xc also because there is resonance in the circuit so when we put the values for omega -not this is 750 radian second multiplied by 0 .8080 henry.

03:12 So this value comes out to be approximately 600 o.

03:18 So both of these reactances are having the same value and the impedance of the circuit will be simply equal to resistance only because xl is equal to xc and z is given by xl minus xc to the whole square plus r squared.

03:38 So as xl cancel out xc so the remaining impedance will be equal to resistance only.

03:48 In the third part of the problem, we have to find the potential drops across various terminals given.

03:56 So first of all we have to find vab and here if we look into the circuit, vab means we have to find potential drop taking place across inductor.

04:08 Vab means vl and that will be given by ib into a circuit...

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