# Find capacitive reactance from Inductive reactance

Jaccobtw
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.

Homework Helper
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What happens to the value for C if the plate spacing is halved? Knowing this you can solve the rest with given information.

• Jaccobtw and Delta2
Homework Helper
Gold Member
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... • Jaccobtw
Jaccobtw
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

• Delta2
Jaccobtw
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... • Delta2
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