Solve Circuit Resonance Homework: 50mA Current & 30mA Frequency Change

In summary: At 50 mA, this impedance is 1V.For 30mA, the impedance would need to be 2/3 the previous value. Since Zl and Zc are proportional to 1/f, we need to find a frequency that is 1/2 that of the previous frequency.In summary, a resonant current of 50mA flowing through a series circuit of a 2mH inductor, 20 ohm resistor, and 0.3nF capacitor has an applied voltage of 1V. When the current is reduced to 30mA, the voltage remains the same but the frequency changes to 1/2 the previous value, corresponding to a phase difference of approximately 53 degrees.
  • #1
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Homework Statement


A resonant current of 50mA flows in a circuit consisting of a 2mH inductor, 20 ohm resistor and 0.3nF capacitor all in series. What is the applied voltage? If the current is reduced to 30mA by changing the frequency but not the voltage, find the new frequency and the phase difference between the voltage and current.

Homework Equations



The j term is the imaginary term in all these equations.

[tex]Z_r = 20[/tex]
[tex]Z_i = j \omega L[/tex]
[tex]Z_c = \frac{1}{j \omega C}[/tex]
[tex]\omega_0 = \frac{1}{\sqrt{LC}}[/tex]

The Attempt at a Solution


Summing the impedances then using Ohm's law I find the voltage to be 1V at 50mA.

When the current is reduced, the voltage is still 1V but the frequency changes. What I've tried is to find the new total impedance using Ohm's law, then I rearrange the total impedance equation I used for the first part to make omega the subject, but I end up with a polynomial and am forced to use an equation to find the roots which I shouldn't need to use and I end up with imaginary terms I don't want. I end up with an answer of 1.3 megaohms, whereas the answer is suppose to be 206kHz (and approx 53 degrees phase).

I'd like some help with this please. I don't know whether I'm along the right lines by trying to find an equation for omega in terms of the component impedances.

Cheers.
 
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  • #2
Never mind, I'm an idiot! I found the value for omega to be 1.3M and assumed this was the frequency, but I didn't divide it by 2 pi which would have given me the correct answer!

Thanks anyway.
 
  • #3
You're making it harder than it is.
Zl = 2PifL
Zc = 1/2PifC.

Total impedance = 20 Ohms + Zl + Zc.
 

1. What is circuit resonance?

Circuit resonance occurs when the inductive and capacitive reactances in a circuit are equal. This leads to a maximum current and voltage in the circuit, resulting in efficient energy transfer.

2. How do you solve circuit resonance problems?

To solve circuit resonance problems, you can use the formula f = 1/(2π√LC), where f is the frequency, L is the inductance, and C is the capacitance. You can also use the graphical method, where you plot the impedance of the circuit against frequency and find the point where it is at its minimum.

3. What is the significance of 50mA current in this homework?

The 50mA current in this homework represents the maximum current in the circuit when it is at resonance. This value is important for calculating the other parameters such as frequency and impedance.

4. Why is a 30mA frequency change important in this homework?

The 30mA frequency change represents a small deviation from the resonant frequency. This allows us to analyze the effects of a change in frequency on the circuit's behavior and understand how it may affect the overall performance.

5. What are some real-life applications of circuit resonance?

Circuit resonance has many practical applications, such as in radio and television broadcasting, where it allows for efficient transfer of electromagnetic waves. It is also used in electronic filters, where it helps to select specific frequencies for signal processing. In addition, resonance is important in electric power systems, where it helps to reduce power losses and improve efficiency.

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