Find capacitive reactance from Inductive reactance

  • Thread starter Thread starter Jaccobtw
  • Start date Start date
  • Tags Tags
    Reactance
Click For Summary
SUMMARY

The discussion focuses on calculating capacitive reactance from inductive reactance in an RLC circuit. The capacitance of a parallel plate capacitor is defined by the formula $$C=\epsilon\frac{S}{d}$$, where halving the plate separation distance $$d$$ results in doubling the capacitance $$C$$. The magnitude of current $$I_0$$ is derived from the equation $$I_0=\frac{V_0}{\sqrt{R^2+(L\omega-\frac{1}{C\omega})^2}}$$, with given values of resistance $$R$$ and inductance $$L$$. The discussion emphasizes setting up equations for the current before and after altering capacitance to solve for the unknowns.

PREREQUISITES
  • Understanding of RLC circuit theory
  • Familiarity with the capacitance formula for parallel plate capacitors
  • Knowledge of current magnitude calculations in AC circuits
  • Ability to manipulate algebraic equations to solve for unknowns
NEXT STEPS
  • Study the derivation of the capacitance formula for parallel plate capacitors
  • Learn about the behavior of RLC circuits under varying capacitance
  • Explore the implications of changing plate separation on capacitance and reactance
  • Investigate the relationship between current, voltage, and impedance in AC circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and capacitive reactance calculations.

Jaccobtw
Messages
163
Reaction score
32
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.
 
Physics news on Phys.org
What happens to the value for C if the plate spacing is halved? Knowing this you can solve the rest with given information.
 
  • Like
Likes   Reactions: Jaccobtw and Delta2
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)
 
  • Like
Likes   Reactions: Jaccobtw
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
 
  • Like
Likes   Reactions: Delta2
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!
 
  • Like
Likes   Reactions: Delta2

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
981
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K