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Q101PE

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College Physics (Urone)
Found in: Page 864

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Short Answer

An RLC series circuit has a 2.50Ω resistor, a 100μH inductor, and a 80.0μF capacitor. (a) Find the circuit's impedance at 120 Hz . (b) Find the circuit's impedance at 5.00 kHz . (c) If the voltage source has Vrms=5.60 V , what is Irms at each frequency? (d) What is the resonant frequency of the circuit? (e) What is Irms at resonance?

  1. The circuit's impedance at 120 Hz is 16.7Ω .
  2. The circuit's impedance at 5 kHz is 3.7Ω .
  3. The Irms at 120 Hz is 0.335 A and at 5 kHz is 1.51 A .
  4. The resonant frequency of the circuit is 1780 Hz .
  5. The Irms at resonance frequency is 2.24 A
See the step by step solution

Step by Step Solution

Step 1: Definition of capacitor

A capacitor is a two-terminal electrical component that may store energy in the form of an electric charge. It is made up of two electrical wires separated by a certain distance. The space between the conductors can be filled with vacuum or a dielectric, which is an insulating substance.

Step 2: Given data

The resistance of the RLC circuit is R=2.50Ω

The capacitance of the RLC circuit is C=80.0μF10-6F1μF=8.00×10-5F

The inductance of the RLC circuit is L=100μH10-6H1μH=1.00×10-4H

Step 3: Write the formula for the capacitor

The reactance of an RLC circuit can be expressed as,

Z=R2+XL-XC2………………(1)

This means we'll need to figure out the capacitor and inductor's active resistances, which can be given as,

XC=12πfC………………….(2)

XL=2πfL…………………..(3)

Step 4: Calculate the circuit's impedance(a)

We have f1=120 Hz, therefore substituting the given data in equations (2) and (3), we get,

XC=12π×120Hz×8.00×10-5F =16.6ΩXL=2π×120Hz×1.00×10-4H =0.075Ω

Now, we use the above-obtained values of XC and XL in equation (3), and we get

Z=2.5Ω2+0.075Ω-16.6Ω2 =16.7Ω

Therefore, the circuit's impedance at 120 Hz is 16.7Ω.

Step 5: Find the circuit's impedance(b)

We have f2=5kHz103Hz1kHz=5×103Hz, therefore substituting the given data in equations (2) and (3), we get,

XC=12π×5×103Hz×8.00×10-5F =0.4ΩXL=2π×5×103Hz×1.00×10-4H =3.14Ω

Now, we use the above-obtained values of XC and XL in equation (3), and we get

Z=2.5Ω2+3.14Ω-0.4Ω2 =3.7Ω

Therefore, the circuit's impedance at 5 kHz is 3.7Ω .

Step 6: Find  at each frequency(c)

According to Ohm's law, the root-mean-square intensity may be calculated using the reactance and the rms potential difference such that,

Irms=VrmsZ…………….(4)

For the frequency 120 Hz , the value of impedance is Z=16.7Ω, therefore

Irms=5.6 V16.7Ω =0.335 A

For the frequency 5 kHz , the value of impedance is Z=3.7Ω, therefore

Irms=5.6 V3.7Ω =1.52 A

Therefore, the Irms at 120 Hz is 0.335 A and at 5 kHz is 1.51 A .

Step 7: Find the resonant frequency of the circuit(d)

The resonant frequency can be calculated using the expression,

f0=12πLC……………(5)

That indicates that in our case, the end outcome will be

f0=12π1.00×10-4H×8.00×10-5F =1780 Hz

Therefore, the resonant frequency of the circuit is 1780 Hz .

Step 8: Find at the resonance frequency(e)

Remember that the resonant frequency occurs when the capacitor and inductor reactance are equal. If we look at the total reactance formula, this suggests that our resistance will be the same as the resistor. To put it another way, the rms current will be,

Irms=5.6 V2.5 Ω =2.24 A

Therefore, the Irms at resonance is 2.24 A .

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