Find capacitive reactance from Inductive reactance

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In summary: That is my secret, I reveal as much as possible without solving the problem, and that is really a lot, and I am confident that it is enough for the OP to solve his problem and learn from it and also not get into trouble with the moderators! :oldbiggrin:In summary, the conversation discusses the relationship between the capacitance of a parallel plate capacitor and the distance between its plates. It is mentioned that the value of capacitance doubles when the plate spacing is halved. The conversation also provides equations for the magnitude of current in an RLC circuit and suggests setting up equations to solve for unknown variables. The summary concludes by mentioning that the conversation provides enough information for the OP to solve the problem without revealing the final solution.
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Homework Statement
In an RLC series circuit that includes a source of alternating current operating at a fixed frequency and voltage, the resistance of 10Ω is equal to the inductive reactance. If the capacitor plate separation is reduced to one half its original value, the magnitude of the current doubles in the circuit. Find the initial capacitive reactance in Ω. Note that although we adjust the separation, you can still treat the capacitor as an ideal parallel plate capacitor.
Relevant Equations
$$Capacitive reactance = X_C = \frac{1}{\omega C} = \frac{V_0}{I_0}$$
$$Inductive Reactance = X_L = \omega L = \frac{V_0}{I_0}$$
I'm just confused about how to go about this problem. I don't have all variables. What does the current doubling with half plate separation have to do with the problem? Your help is much appreciated. Thank you.
 
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  • #2
What happens to the value for C if the plate spacing is halved? Knowing this you can solve the rest with given information.
 
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Yes exactly as @hutchphd said. The capacitance of an ideal parallel plate capacitor is given by $$C=\epsilon\frac{S}{d}$$ where ##S## the surface of one of the plates and ##d## their separation distance. So if ##d## gets halved, then ##C## gets ...

You will also need the standard equation of RLC circuit for the magnitude of current ##I_0##, $$I_0=\frac{V_0}{\sqrt{R^2+(L\omega-\frac{1}{C\omega})^2}}$$

It is given that ##R=L\omega=10\Omega##. I suggest you set $$x=\frac{1}{C\omega}$$ as the unknown cause that's what the problem is asking for (I know ##\omega## and ##C## are both unknowns but thankfully the problem doesn't ask for each of them, only for the x.

Then set $$x'=\frac{1}{C'\omega}$$ where ##C'## the capacitance after the distance between the plates get halved.

Find the equation between ##x## and ##x'## (This is really easy)

Then find ##I_0(x)## and ##I_0(x')## that is the magnitude of current before and after the altering of capacitance.

Find the equation between ##I_0(x)## and ##I_0(x')## (This is also quite easily deduced from the statement of the problem)

You 'll have two equations and two unknowns ##x## and ##x'##

I think I really helped a lot, a moderator might see this and edit my post...:nb)
 
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hutchphd said:
What happens to the value for C if the plate spacing is halved? Knowing this you can solve the rest with given information.
The value of C doubles
 
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Delta2 said:
Yes exactly as @hutchphd said. The capacitance of an ideal parallel plate capacitor is given by $$C=\epsilon\frac{S}{d}$$ where ##S## the surface of one of the plates and ##d## their separation distance. So if ##d## gets halved, then ##C## gets ...

You will also need the standard equation of RLC circuit for the magnitude of current ##I_0##, $$I_0=\frac{V_0}{\sqrt{R^2+(L\omega-\frac{1}{C\omega})^2}}$$

It is given that ##R=L\omega=10\Omega##. I suggest you set $$x=\frac{1}{C\omega}$$ as the unknown cause that's what the problem is asking for (I know ##\omega## and ##C## are both unknowns but thankfully the problem doesn't ask for each of them, only for the x.

Then set $$x'=\frac{1}{C'\omega}$$ where ##C'## the capacitance after the distance between the plates get halved.

Find the equation between ##x## and ##x'## (This is really easy)

Then find ##I_0(x)## and ##I_0(x')## that is the magnitude of current before and after the altering of capacitance.

Find the equation between ##I_0(x)## and ##I_0(x')## (This is also quite easily deduced from the statement of the problem)

You 'll have two equations and two unknowns ##x## and ##x'##

I think I really helped a lot, a moderator might see this and edit my post...:nb)
Cool! Thanks for your help!
 
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FAQ: Find capacitive reactance from Inductive reactance

What is capacitive reactance and inductive reactance?

Capacitive reactance is the opposition to the flow of alternating current (AC) caused by a capacitor. Inductive reactance is the opposition to the flow of AC caused by an inductor.

Why is it important to find the capacitive reactance from inductive reactance?

Knowing the capacitive reactance from inductive reactance is important in understanding the behavior of AC circuits and designing them for specific purposes.

How do you calculate the capacitive reactance from inductive reactance?

The formula for finding capacitive reactance from inductive reactance is Xc = 1/(2πfC), where Xc is the capacitive reactance, f is the frequency of the AC, and C is the capacitance.

What are the units for capacitive reactance and inductive reactance?

The units for both capacitive and inductive reactance are ohms (Ω).

What are the practical applications of finding capacitive reactance from inductive reactance?

Some practical applications include designing filters for electronic circuits, improving power factor in AC systems, and controlling the flow of current in motors and generators.

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