Calculating RMS Currents and Power in a High-Pass Filter Circuit

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In summary, the circuit consists of a load resistor with RL = 20.6 Ω connected to a high-pass filter with an inductor of L = 3.20-mH and a resistor of R = 4.00-Ω. The output of the ideal ac generator is given by = (100 V) cos(2πft). We need to find the rms currents in all three branches of the circuit for a driving frequency of 475 Hz. The fraction of total average power supplied by the ac generator that is delivered to the load resistor needs to be determined for both 475 Hz and 1840 Hz. To solve this problem, we can use equations for impedance and current, where z = sqrt(R^
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Homework Statement


The figure below shows a load resistor that has a resistance of RL = 20.6 Ω connected to a high-pass filter consisting of an inductor that has inductance L = 3.20-mH and a resistor that has resistance R = 4.00-Ω. The output of the ideal ac generator is given by = (100 V) cos(2πft).
Image at: http://www.webassign.net/tipler6/29-p-041.gif
upload_2016-2-7_3-20-52.png
(note: image inserted by mentor)

Find the rms currents in all three branches of the circuit if the driving frequency is 475 Hz.
For Irms, I(Rl)rms, I(L)rms

Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 475 Hz.

Find the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 1840 Hz.

Homework Equations


z= sqrt(R^2 +(wL-1/wC)2)
I=Vsource/Ztotal
I = Imax*cos(2*pi*f*t-theta)

The Attempt at a Solution



So I am in a low level physics and we got stuck with a problem that my TA has told me he hasn't had to solve until about his senior year. He doesn't really understand why we have this problem. This is way over my head and I will need a lot of help if anyone is willing to give it. If anyone would like to help (which I would highly appreciate) please talk to me like I'm dumb because I am VERY confused. Thanks.

Combining the inductor and load resistor we get a z equation of z=R+ (Rl(wL)^2/(Rl^2+(wL)^2)) + Rl^2*w*L/(Rl^2+(wL)^2)i where i means an imaginary number

Then |Z|total= sqrt([Zreal]^2+[Zimaginary]^2)

Then I = Vsource/Ztotal where Vsource=Vrms=100/sqrt(2)
Then I = Imax*cos(2*pi*f*t - theta)

I attempted and got |Z|= 8.47 and I = 8.35A which I thought would be Irms but I was wrong
 
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I(Rl)rms= |Vsource-I(L)|/RLI(L)rms= I(Rl)rms*wL*sqrt(1/(RL^2+(wL)^2))For the power, P= Vsource*IFraction of total average power delivered to load resistor = Pload/PtotalI'm totally lost and would really appreciate help. Thanks.
 

1. How do I calculate the value of Irms in a circuit?

To calculate Irms, you need to find the root mean square (RMS) value of the current in the circuit. This can be done by taking the square root of the average of the squared values of the current over one complete cycle. Alternatively, you can use a multimeter or an oscilloscope to directly measure the RMS value of the current.

2. Why is it important to find the value of Irms in a circuit?

Irms is the effective or average current in a circuit. It takes into account both the magnitude and direction of the current, unlike peak or maximum current values. This is important because many electronic components, such as resistors and capacitors, are rated for their ability to handle RMS current. Therefore, knowing the Irms value can help ensure the safe and efficient operation of a circuit.

3. Can Irms be higher than the maximum current in a circuit?

Yes, it is possible for Irms to be higher than the maximum current in a circuit. This can occur when the current is not a perfect sinusoidal waveform, which is often the case in real-world circuits. In these situations, the RMS value of the current will be higher than the maximum value.

4. How does the frequency of the current affect the value of Irms?

The frequency of the current does not directly affect the value of Irms. However, it can indirectly impact it by affecting the shape of the current waveform. For example, a higher frequency can result in a higher RMS value if the current waveform is distorted.

5. Can I use the same method to find Irms in AC and DC circuits?

No, the methods for finding Irms in AC and DC circuits are different. In a DC circuit, the Irms value is equal to the average current. However, in an AC circuit, the Irms value must be calculated using the RMS method described in the first question. Additionally, the instruments used to measure Irms may also differ between AC and DC circuits.

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