If a higher value capacitor is installed, what happens to the voltage/current relationship?

Thread Starter

tylersmith66674

Joined Jun 8, 2022
7
I am stumped, I am aware there is a calculation to describe this, or one can describe it by analysing the back emf and magenetic field, but I am not sure about either.
What would happen to the relationship?
 

Papabravo

Joined Feb 24, 2006
21,261
So, have you calculated the inductive reactance in ohms (Ω) for the inductor? Do you know the expression for capacitive reactance? It is a function of frequency and capacitance.

BTW -- what is CIVIL?

Have you ever heard of the mnemonic: "ELI the ICE man"? It is a handy way to remember the relationship you are after.
https://www.electrical4u.com/eli-the-ice-man/
 
Last edited:

Thread Starter

tylersmith66674

Joined Jun 8, 2022
7
So, have you calculated the inductive reactance in ohms (Ω) for the inductor? Do you know the expression for capacitive reactance? It is a function of frequency and capacitance.

BTW -- what is CIVIL?

Have you ever heard of the mnemonic: "ELI the ICE man"? It is a handy way to remember the relationship you are after.
https://www.electrical4u.com/eli-the-ice-man/
I have calculated the size of the capacitor to be 239 micro farads
 

Thread Starter

tylersmith66674

Joined Jun 8, 2022
7
I’m not sure, what would happen to the voltage / current phase relationship if I increased the size of the capacitor ?
The voltage would lag the current ?
 

Papabravo

Joined Feb 24, 2006
21,261
Focus on one thing at a time.
  1. If you have 2 impedances in parallel and one of them is a capacitor, and you double the value of the capacitor. What happens to the impedance of the capacitor leg of the parallel circuit?
  2. If you have a parallel circuit with two equal impedances and you make one of them bigger or smaller, what happens to the voltage and current?
 

Papabravo

Joined Feb 24, 2006
21,261
I’m not sure if I’m perfectly honest
OK, this is Homework Help, not homework done for you. You have computed the value of a capacitor to have the same impedance as an inductor, so presumably you can compute the impedance of a capacitor that is twice the size. Hint: it gets smaller

If you have the parallel combination of Z1 || Z2, and Z1 is equal to Z2, then the combined impedance is
\( \cfrac{Z\times Z}{Z+ Z}\;=\;\cfrac{Z^2}{2Z}\;=\;\cfrac{Z}{2} \)
Now replace one of the impedances with a value that is 2 times bigger
\( \cfrac{Z\times 2Z}{Z+2Z}\;=\;\cfrac{2Z^2}{3Z}\;=\;\cfrac{2}{3}Z \)
Notice that a parallel combination of impedances will always have a total that is less than either of the components.
Redo the above exercise except make one of the impedances be Z/2 instead of 2Z.
Does that help?
 
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