Heat dissipated in a Resistor-Capacitor circuit

In summary: Volts? ##C_1## started out at 4 V and dropped to 3 V. How did ##C_1## manage to drive current "uphill"?This looks like a differential equation to me. Is this the kind of class where you might be expected to solve a differential equation?I think if it is a class where you are given the formula for RC capacitor discharge with a time constant (we did this in High School Physics) then you don't need to know differential equations to find an expression for voltage across the resistor at a particular time. But it might be a little more complicated. hmmm.
  • #1
Prabs3257
65
4
Homework Statement
A charged capacitor c1= 12uF with voltage 4 volt is connected to uncharged capacitor c2=4uF and a resistance of 4 ohm as shown in the diagram . total heat generated across resistance is H upto a time interval t0 if at t=t0 voltage across C1 is 3V then value of H/4 is
Relevant Equations
Energy conservation
I tried to conserve the charge on the left plates of both the capacitors as intially the total charge on both is 48 and at t=t0 the total charge is 36(on c1) +4V(V is the potential across c2) so i got V=3 and then i conserved the energy
Initial energy on both capacitor = final energy on both + heat due to resistor but then i got H = 26 but as per the answer given H should be 24 please tell me where am i wrong
 

Attachments

  • IMG_20200307_234225.jpg
    IMG_20200307_234225.jpg
    62.9 KB · Views: 151
Last edited by a moderator:
Physics news on Phys.org
  • #2
Prabs3257 said:
Homework Statement:: A charged capacitor c1= 12uF with voltage 4 volt is connected to uncharged capacitor c2=4uF and a resistance of 4 ohm as shown in the diagram . total heat generated across resistance is H upto a time interval t0 if at t=t0 voltage across C1 is 3V then value of H/4 is
Relevant Equations:: Energy conservation

I tried to conserve the charge on the left plates of both the capacitors as intially the total charge on both is 48 and at t=t0 the total charge is 36(on c1) +4V(V is the potential across c2) so i got V=3 and then i conserved the energy
How do you arrive at the potential across ##C_2## is 4 Volts? ##C_1## started out at 4 V and dropped to 3 V. How did ##C_1## manage to drive current "uphill"?
 
  • #3
Hmmm ... ##t_0## is an arbitrary point in time, not some equilibrium point at infinite time that you might know something about. This looks like a differential equation to me. Is this the kind of class where you might be expected to solve a differential equation?
 
  • #4
I think if it is a class where you are given the formula for RC capacitor discharge with a time constant (we did this in High School Physics) then you don't need to know differential equations to find an expression for voltage across the resistor at a particular time. But it might be a little more complicated. hmmm. I am on my phone at the moment. Let me think about it.
 
  • #5
OK, I don't think you need any differential equations, or even the time domain expression of voltage. Take a look at this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

Note that charge is conserved, and the resistor does not store any charge. All charge that leaves the first capacitor, goes to the 2nd capacitor. You can use the online interactive calculator (or the formulas) and find the amount of charge and amount of energy in the first capacitor, initially. Then plug in the new voltage and find the amount of charge and energy left when it is 3 volts. Then you can use the formulas, to find out how much energy the 2nd capacitor has, based on its charge and capacitance. You can also calculate its voltage.

So you have energy left the first capacitor, and the second capacitor has some energy on it. The difference must be what was dissipated as heat.
 
  • Like
Likes TSny
  • #6
scottdave said:
OK, I don't think you need any differential equations, or even the time domain expression of voltage. Take a look at this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

Note that charge is conserved, and the resistor does not store any charge. All charge that leaves the first capacitor, goes to the 2nd capacitor. You can use the online interactive calculator (or the formulas) and find the amount of charge and amount of energy in the first capacitor, initially. Then plug in the new voltage and find the amount of charge and energy left when it is 3 volts. Then you can use the formulas, to find out how much energy the 2nd capacitor has, based on its charge and capacitance. You can also calculate its voltage.

So you have energy left the first capacitor, and the second capacitor has some energy on it. The difference must be what was dissipated as heat.
Ya what i did was something along the same line i conserved thw charge between the 2 capacitor plates as intially there was total of 48 uC charge so at t0 we were given the charge on c1(q=cv where v was given to us as 3) so at t0 it became 36 + charge on c2 at t0 equating initial and final i got the potential of c2 at t0 hence the initial and final energy
 
  • #7
Prabs3257 said:
Ya what i did was something along the same line i conserved thw charge between the 2 capacitor plates as intially there was total of 48 uC charge so at t0 we were given the charge on c1(q=cv where v was given to us as 3) so at t0 it became 36 + charge on c2 at t0 equating initial and final i got the potential of c2 at t0 hence the initial and final energy
But the ans i am getting is not correct as per the answer key
 
  • #8
gneill said:
How do you arrive at the potential across ##C_2## is 4 Volts? ##C_1## started out at 4 V and dropped to 3 V. How did ##C_1## manage to drive current "uphill"?
gneill said:
How do you arrive at the potential across ##C_2## is 4 Volts? ##C_1## started out at 4 V and dropped to 3 V. How did ##C_1## manage to drive current "uphill"?
potential across c2 is not 4 the 4V i wrote represented the charge (capcitance(4) * potential(V)) you must have thought the V is the si unit of potential but actually V Is the potential of c2 at t0 sorry about the formating
 
  • #9
When I did all of those calculations, I got 24 μJ energy difference between C1 and C2 energies.

Can you show your calculations, and we can help you figure out what happened?
 
  • #10
ohh no really sorry i did the calculations again and got the correct answer just did a multiplication error earlier :sorry: i thought i was conceptually wrong really sorry for all the trouble to everyone and thanks a lot for your help
 

Related to Heat dissipated in a Resistor-Capacitor circuit

1. What is a resistor-capacitor circuit?

A resistor-capacitor circuit is an electrical circuit that contains both a resistor and a capacitor. The resistor limits the flow of current, while the capacitor stores electrical charge. Together, they can be used to control the flow of electricity in a circuit.

2. How does heat dissipate in a resistor-capacitor circuit?

Heat is dissipated in a resistor-capacitor circuit through a process called Joule heating. This occurs when the electric current flowing through the resistor encounters resistance, causing the electrons to collide with the atoms in the resistor and transfer their kinetic energy into heat.

3. What factors affect the amount of heat dissipated in a resistor-capacitor circuit?

The amount of heat dissipated in a resistor-capacitor circuit is affected by the resistance of the resistor, the capacitance of the capacitor, and the amount of current flowing through the circuit. The higher the resistance and capacitance, and the greater the current, the more heat will be dissipated.

4. How can heat dissipation be minimized in a resistor-capacitor circuit?

To minimize heat dissipation in a resistor-capacitor circuit, the resistance and capacitance can be decreased, and the current can be reduced. Additionally, using larger resistors and capacitors can help to disperse the heat more effectively.

5. What are the potential risks of excessive heat dissipation in a resistor-capacitor circuit?

Excessive heat dissipation in a resistor-capacitor circuit can lead to overheating, which can damage or destroy the components in the circuit. It can also cause a decrease in the circuit's efficiency and potentially create a fire hazard. It is important to properly design and manage heat dissipation in these circuits to prevent any potential risks.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
457
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
305
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
2
Replies
62
Views
3K
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top