Difficult Capacitor problem

  • Thread starter Hashiramasenju
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In summary, the conversation discusses a circuit with a slide-wire of uniform resistance and a moveable contact. The total length of the slide-wire, voltage, and capacitance are known, and the goal is to find the energy dissipated. The switches in the circuit charge and discharge the capacitor, and the voltage across the capacitor is not initially 24V. The slide-wire also has significant resistance, which should be taken into account when calculating energy.
  • #1
Hashiramasenju
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Homework Statement


https://isaacphysics.org/api/images/content/questions/physics/circuits/capacitors/level4/figures/Circuits_potentiometer_capacitor_otp_2.svg
PQ is a slide-wire of uniform resistance, and J is a moveable contact at some point along it, such that the length from P to J is a known distance l. The total length of PQ is also known to be 1.00m. V=24.0V, C=10.0mF.
Find the energy dissipated. Given that initially S1 is closed and S2 is open, and l=326mm. The circuit is left in this arrangement for long enough that the circuit reaches a steady state. Then S2 is opened and S1 is closed simultaneously.

Homework Equations


E=0.5cv^2

The Attempt at a Solution


So using the above eqn i assumed that the pd across the capacitor will be 24v(because the same voltage flows through the variable resistor branch and the capacitor branch. Now as the current stablies there should be no current flowing through the capacitor. Hence the pd across the capacitor is 24v. Now the total energy of the capacitor = total energy dissipated by the resistor ) then i calculated the energy E=0.5*10*(10^-3)*24^2=2.88
Is this right baecause the answer is different
 
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  • #2
Hashiramasenju said:
Find the energy dissipated. Given that
initially S is closed and S is open
Please include the subscripts on the switch labels; i.e., the S's.

Think a bit more about whether or not the pd across the capacitor is 24V initially.
 
  • #3
What do you think is achieved by those switches? Nothing?
 
  • #4
NascentOxygen said:
What do you think is achieved by those switches? Nothing?
S1 basically charges the capacitor but S2 helps discharge the capacitor and time constant =RC=R*10*10^-3
I have a doubt that the pd across the capacitor is not 24v.
Otherwise the energy dissipated in the resistor is not equal to the energy the capacitor posseses
 
  • #5
Hashiramasenju said:
I have a doubt that the pd across the capacitor is not 24v.
PD across capacitor is never even close to 24v here. What is the function of the slider on the resistance wire?

Otherwise the energy dissipated in the resistor is not equal to the energy the capacitor posseses
Does the question ask for the energy dissipated in specifically resistor R?
 
  • #6
NascentOxygen said:
PD across capacitor is never even close to 24v here. What is the function of the slider on the resistance wire?Does the question ask for the energy dissipated in specifically resistor R?
But shouldn't the voltage across a parallel circuit be the same regardless of the resistance?
 
  • #7
Line PJQ is a slide resistor. So section PJ has significant resistance. Likewise for section JQ. Might be helpful if you included resistor symbols in these two sections.
 
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  • #8
Thanks a lot got it
 

1. What is a "Difficult Capacitor problem"?

A "Difficult Capacitor problem" is a type of problem that involves calculating the capacitance or charge in a circuit with multiple capacitors connected in a complex manner. These problems often require the use of advanced mathematical techniques and careful analysis of the circuit.

2. How do I approach solving a "Difficult Capacitor problem"?

The key to solving a "Difficult Capacitor problem" is to break down the circuit into simpler components and then use the appropriate equations to calculate the overall capacitance or charge. It is also important to carefully label all the components and keep track of the units to ensure an accurate solution.

3. What are some common techniques used to solve "Difficult Capacitor problems"?

Some common techniques used to solve "Difficult Capacitor problems" include the use of the equivalent capacitance formula for capacitors in series and parallel, the use of Kirchhoff's laws, and the application of the principle of conservation of charge. Other techniques such as simplifying the circuit and using symmetry can also be helpful.

4. What are some challenges that may arise when solving a "Difficult Capacitor problem"?

One of the main challenges when solving a "Difficult Capacitor problem" is keeping track of all the components and their values, as well as the units. Another challenge may arise when dealing with non-ideal components or when the circuit is significantly complex, making it difficult to apply the equations directly.

5. Are there any tips for solving "Difficult Capacitor problems" more efficiently?

One tip for solving "Difficult Capacitor problems" more efficiently is to thoroughly understand the concepts and equations involved, and to practice with simpler problems before tackling more complex ones. It is also helpful to double-check all calculations and to use multiple methods to arrive at the solution for verification.

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