Calculating Energy Dissipation in an RC Circuit

In summary, the question asks for the amount of energy dissipated by a resistor in a circuit with a 50V battery, 125 ohm resistor, and 20uF capacitor. The switch is flipped from position a to b for 1.25 ms and then back to a. To solve this problem, you need to find the total charge in the first capacitor and the charge transferred to the second capacitor in 1.25 ms. Then, using the average current and the power dissipation formula, you can calculate the energy dissipated by the resistor by integrating the average current squared. The answer is approximately 23 mJ.
  • #1
ursulan
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Homework Statement


The switch in the figure (Intro 1 figure) has been in position a for a very long time. It is suddenly flipped to position b for 1.25 ms, then back to a. How much energy is dissipated by the resistor?

the circuit for a is 50V battery, 125 ohm resistor and 20uF capacitor. The circuit for b is the 20uF capacitor and the 50 ohm resistor.

Homework Equations



Time constant=RC, Ucap=0.5CV^2, V=IR, Q=CV, i don't know I've tried others too

The Attempt at a Solution



I know the capacitor will be at 50V. I found time constant and current and charge and tried so see if i=i(max)e^(-t/RC) would work and tried to relate it back to energy but it didn't. The answer is 23 mJ.
 
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  • #2
First of all find the total charge in the first capacitor when the switch in position a.
Find the total charge transferred to second capacitor in 1.25 ms when the switch is in b position. dQ/dt will give you the current in the resistance. Find the average current I and I^2R will give you the power dissipation.
 
Last edited:
  • #3
To get answer integrate I^2R dt. Integrating the average current squared is an approximation...
 

1. What is an RC circuit?

An RC circuit is an electrical circuit that contains a resistor (R) and a capacitor (C). The resistor limits the flow of electrical current, while the capacitor stores electrical energy.

2. How does an RC circuit work?

When a voltage source is connected to an RC circuit, the capacitor starts to charge. As it charges, it creates an opposing voltage that reduces the flow of current. Once the capacitor is fully charged, no current flows through the circuit. When the voltage source is removed, the capacitor discharges and releases its stored energy.

3. What is the time constant of an RC circuit?

The time constant of an RC circuit is the amount of time it takes for the capacitor to charge or discharge to 63.2% of its maximum charge or discharge. It is calculated by multiplying the resistance (R) by the capacitance (C) of the circuit.

4. How does the energy stored in a capacitor change in an RC circuit?

In an RC circuit, the energy stored in the capacitor initially increases as it charges and reaches a maximum value when the capacitor is fully charged. As the capacitor discharges, the energy stored decreases until it reaches zero when the capacitor is fully discharged.

5. What is the relationship between the time constant and the energy stored in an RC circuit?

The time constant and the energy stored in an RC circuit are inversely proportional. As the time constant increases, the energy stored decreases. This means that a larger resistance or capacitance will result in a longer time constant and a smaller amount of energy stored in the capacitor.

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