Energy in capacitor at steady state

In summary, the conversation discusses the steady state energy stored in two capacitors in a circuit. The equations used are Kirchoff's laws, E=0.5CV^2, and V=IR. The conversation concludes that the capacitors can be effectively considered in series, and the currents in the upper and lower loops will be isolated and fixed at steady state.
  • #1
Krushnaraj Pandya
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Homework Statement


In the circuit shown, C1= 1 microfarad, C2=3 microfarad, in steady state, the energy stored in these capacitors are?

Homework Equations


Kirchoff's laws, E=0.5CV^2, V=IR

The Attempt at a Solution


At steady state, no current passes through the capacitors, so current is isolated in upper and lower loops- Using KVL in them I(upper)=1 Ampere and I(lower)=0.5 Ampere, Then using KVL in loops with both capacitors gives 8+V1+V2=0 where V1,V2 are potential across capacitors. I don't know how to proceed further, I'd be grateful for your help- thank you.
 

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  • #2
So effectively you have 8 V across two capacitors that are in series...
 
  • #3
gneill said:
So effectively you have 8 V across two capacitors that are in series...
the same current isn't flowing through them, in fact no current is flowing through them at all- how can we say they're in series?
 
  • #4
gneill said:
So effectively you have 8 V across two capacitors that are in series...
I did get the correct answers using that assumption...so now the only thing I want to understand is how we can consider them in series
 
  • #5
Krushnaraj Pandya said:
the same current isn't flowing through them, in fact no current is flowing through them at all- how can we say they're in series?
Krushnaraj Pandya said:
I did get the correct answers using that assumption...so now the only thing I want to understand is how we can consider them in series
Note that I said "effectively". By your own KVL you wrote: 8+V1+V2=0. That can be interpreted as effectively a series connection.
 
  • #6
gneill said:
Note that I said "effectively". By your own KVL you wrote: 8+V1+V2=0. That can be interpreted as effectively a series connection.
I don't think I grasped it, so any loop we take in KVL, we can assume all the capacitors along the way to be in series?
 
  • #7
Krushnaraj Pandya said:
I don't think I grasped it, so any loop we take in KVL, we can assume all the capacitors along the way to be in series?
I wouldn't go that far. But writing KVL at steady state you're just looking at fixed potential changes with no current flowing.

Also, if you look carefully at the circuit, you'll note that the currents through the capacitors must be the same at all times. Any current that wants to get to the bottom loop via one capacitor must return to the top loop via the other, and vice versa.
 
  • #8
gneill said:
But writing KVL at steady state you're just looking at fixed potential changes with no current flowing.
What about the current flowing through the parts of the loop besides the capacitor branches? They aren't fixed potential changes.

gneill said:
Also, if you look carefully at the circuit, you'll note that the currents through the capacitors must be the same at all times. Any current that wants to get to the bottom loop via one capacitor must return to the top loop via the other, and vice versa.
Ah! that's very reasonable and intuitive, Thank you.
 
  • #9
Krushnaraj Pandya said:
What about the current flowing through the parts of the loop besides the capacitor branches? They aren't fixed potential changes.
At steady state they will be. The currents in the upper and lower loops will be isolated and fixed.
 
  • #10
gneill said:
At steady state they will be. The currents in the upper and lower loops will be isolated and fixed.
Alright, I understand, Thank you very much for your help :D
 
  • #11
You're welcome.
 
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Related to Energy in capacitor at steady state

1. What is the definition of steady state in a capacitor?

The steady state in a capacitor refers to the condition where the flow of electric charge has reached a constant rate, meaning the voltage and current across the capacitor remain constant over time.

2. How is energy stored in a capacitor at steady state?

At steady state, the energy in a capacitor is stored in the electric field between the two plates. As charge builds up on one plate, an electric field is created, and energy is stored in this field. The amount of energy stored is proportional to the capacitance and the square of the voltage across the capacitor.

3. What happens to the energy in a capacitor if the voltage is increased at steady state?

If the voltage across a capacitor is increased at steady state, the energy stored in the capacitor will also increase. This is because the electric field between the plates will become stronger, and more energy will be stored in this field.

4. Can energy be lost in a capacitor at steady state?

No, energy cannot be lost in a capacitor at steady state. The energy is constantly being exchanged between the electric field and the charges on the plates, but the total amount of energy remains constant.

5. How does the energy in a capacitor at steady state compare to the energy in an inductor at steady state?

The energy in a capacitor at steady state is stored in the electric field, while the energy in an inductor at steady state is stored in the magnetic field. Both types of energy are constantly exchanged between the field and the charges or current, but the total amount of energy remains constant in both cases.

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