# RC circuit and energy

• ursulan
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.

## 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.

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.

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To get answer integrate I^2R dt. Integrating the average current squared is an approximation...