Impedance of practical Capacitor

In summary, the impedance of a practical capacitor is a measure of the resistance to the flow of alternating current through the capacitor. It is affected by the frequency of the current, the physical properties of the capacitor, and the surrounding circuit. The impedance of a capacitor is inversely proportional to the frequency, and can be calculated using the capacitance and the frequency. In practical applications, capacitors with higher impedance are preferred as they provide better filtering and stability in electronic circuits.
  • #1
bigu01
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0

Homework Statement


A practical capacitor can be modeled by an ideal capacitor in parallel with a resistor.Find the impedance of practical capacitor at the radian frequency ω=377rad/s.Known C1=0.1 x 10^-6F R1=1 *10^6.

Homework Equations



1/Z=1/R1 + jCω

The Attempt at a Solution

I am first determining the Z, by adding the impedance of resistor and the capacitor in parallel.
And I am getting Z1(377)=10^6/(1+j37.7)= 2.6516*10^4 ∠-1.519 However book solutions gives ∠-1.5443.I would like to get some help about where my mistake is,or how should I approach the question. I know that finding the angle we should do θ=arctan(y/x) where y is the imaginary part and x is the real part.Hope I was clear and did not complicate it.
 
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  • #2
Please show your reasoning and working.
 
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  • #3
R//C Z=R//1/jωC = R1/(1+jωC1R1) inputing numbers I get =10^6/(1+j37.7) multiplying by 1+j37.7's conjugate I got 703.09-26506.54j , tried to find the length got the 26515.86 , finding the angle theta=arctan(-26506.54/703.09) doesn't give me same as book result. I get -1.519 book got -1.5443
 
  • #4
Looking at my procedure everything looks fine.I am wondering if book has done any mistake,or there is something else I should do, which I have missed,and it is making me do the mistake.
 
  • #5
bigu01 said:
R//C Z=R//1/jωC = R1/(1+jωC1R1) inputing numbers I get =10^6/(1+j37.7) multiplying by 1+j37.7's conjugate I got 703.09-26506.54j , tried to find the length got the 26515.86 , finding the angle theta=arctan(-26506.54/703.09) doesn't give me same as book result. I get -1.519 book got -1.5443

Your numbers all look fine except for the result of the arctan(). When I enter that argument and take the arctan I get the book's result.
 
  • #6
gneill said:
Your numbers all look fine except for the result of the arctan(). When I enter that argument and take the arctan I get the book's result.

You're saying that arctan(26506.54/703.09) gives 1.554 , either my two calculators are wrong or I am doing something wrong.
 
  • #7
bigu01 said:
You're saying that arctan(26506.54/703.09) gives 1.554 , either my two calculators are wrong or I am doing something wrong.

That's what I'm saying :smile:
 
  • #8
Okay then, what about this one 101.92+j90.38,what do you get as your ∠? The book answer is ∠0.723 and about this one I have no clue how did they get it
 
  • #9
bigu01 said:
Okay then, what about this one 101.92+j90.38,what do you get as your ∠? The book answer is ∠0.723 and about this one I have no clue how did they get it

The numbers are probably rounded somewhat. I get 0.725 radians as the angle.
 
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  • #10
Multiple threads are confusing... it looks like the error was here:
bigu01 said:
I was using the degree mode, I should have used the radian angle unit one

I have no idea how you get -1.519 in degree mode... well does not matter.
 

FAQ: Impedance of practical Capacitor

1. What is the definition of impedance for a practical capacitor?

The impedance of a practical capacitor is the measurement of the opposition to the flow of alternating current (AC) through the capacitor. It is represented by the symbol Z and is measured in ohms. It takes into account both the capacitive reactance and the resistance of the capacitor.

2. How is the impedance of a practical capacitor calculated?

The impedance of a practical capacitor can be calculated using the formula Z = 1/(jωC), where j is the imaginary unit, ω is the frequency of the AC signal, and C is the capacitance of the capacitor. This formula takes into account the capacitive reactance, which is inversely proportional to the frequency, and the resistance of the capacitor.

3. What factors affect the impedance of a practical capacitor?

The impedance of a practical capacitor is affected by several factors, including frequency, capacitance, and temperature. As the frequency of the AC signal increases, the capacitive reactance decreases, resulting in a decrease in impedance. Similarly, as the capacitance of the capacitor increases, the impedance decreases. However, an increase in temperature can lead to an increase in the resistance of the capacitor, resulting in an increase in impedance.

4. How does the impedance of a practical capacitor affect its performance in a circuit?

The impedance of a practical capacitor plays a crucial role in its performance in a circuit. It determines the amount of current that can flow through the capacitor and affects the phase relationship between the current and voltage. A lower impedance allows for a higher current flow and a higher capacitance, resulting in a better power factor and improved circuit efficiency.

5. Can the impedance of a practical capacitor be reduced?

Yes, the impedance of a practical capacitor can be reduced by increasing the frequency of the AC signal, increasing the capacitance of the capacitor, or using a capacitor with a lower temperature coefficient of resistance. Additionally, using capacitors in parallel can also decrease the overall impedance. However, it is important to choose the appropriate capacitor for the specific circuit and application to ensure optimal performance and reliability.

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