∙In an R-L-C series circuit, R=300 Ω, L=0.400 H, and C=6.00 ×10^-8 F . When the ac source operat

step 1 For Part A, we want the voltage amplitude. And we're going to get that by setting it equal to I

∙In an R-L-C series circuit, R=300 Ω, L=0.400 H, and C=6.00...

$\bullet$ In an $R-L-C$ series circuit, $R=300 \Omega, L=0.400 \mathrm{H},$ and $C=6.00 \times 10^{-8} \mathrm{F}$ . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 $\mathrm{A}$ . (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

$\bullet$ In an $R-L-C$ series circuit, $R=300 \Omega, L=0.400 \mathrm{H},$ and
$C=6.00 \times 10^{-8} \mathrm{F}$ . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 $\mathrm{A}$ . (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Key Concepts

Average Power in AC Circuits

Average power in AC circuits is typically determined using the root-mean-square (rms) values of voltage and current and considering the phase difference between them. In a circuit where the voltage and current are in phase—as is the case at resonance with a purely resistive impedance—the average power is simply the product of the rms voltage and rms current. This concept is crucial for analyzing how much energy is dissipated as heat in the resistor or used by the load.

Voltage Amplitude and Phase Relationships

Voltage amplitude in AC circuits refers to the maximum value of the sinusoidal voltage waveform. In analyzing series circuits, understanding how the voltage divides among components (resistor, inductor, and capacitor) is crucial, especially considering their phase relationships. At resonance, while the source voltage appears across the resistor with no phase shift, the voltages across the inductor and capacitor can be significantly higher due to reactive effects, although they are 180 degrees out of phase with each other.

Impedance in AC Circuits

Impedance is the generalized resistance in an AC circuit that includes both the real resistance and the imaginary reactance components. It determines how the circuit opposes the flow of alternating current, capturing the effects of inductors and capacitors whose reactances vary with frequency.

Resonance in LRC Circuits

Resonance in a series LRC circuit occurs when the inductive reactance equals the capacitive reactance, causing them to cancel each other out. This results in the total impedance being purely resistive. At resonance, the circuit current reaches its maximum possible value for a given voltage, and the voltage across the resistor is in phase with the current.

Transcript

00:01 For part a, we want the voltage amplitude.

00:03 And we're going to get that by setting it equal to i times the impedance, where i is the current amplitude.

00:12 Since we're at resonance, this is equal to ir, since the impedance is equal to the resistance at resonance.

00:20 And then now we can plug in the things we know.

00:22 We know the current amplitude is 0 .5 amps, and the r, the resistance, is 300 oms.

00:29 And so this gives 150 volts for v.

00:39 For b, we need three things.

00:41 We need the voltage across the resistor, which is equal to i times the resistor.

00:49 But that's just going to be this voltage here.

00:51 So it's also 150 volts.

00:55 To figure out what the voltage is across the inductor and capacitor, we have to figure out what the reactances of those components are.

01:03 So the reactance of the inductor is omega -l, which is equal to l times omega when we're at resonance, which is 1 over square root of l times the capacitance.

01:18 This is what omega is when we're at the resonant frequency, which is what we're at in this problem.

01:27 So plugging in everything, we get 2 ,582 oms.

01:32 And then now that we have the reactants is, we can figure out what the voltage is across the conductor.

01:38 So it's equal to the current amplitude.

01:40 Times the reactants.

01:43 And so this gives 1 ,290 volts.

01:48 Same thing for the capacitor.

01:50 This is equal to 1 over omega -c, but we know what omega is...