# Energy dissipated through resistor

• hopkinmn
In summary, the student attempted to solve a problem involving power and current, but was having difficulty with the equations. They were able to find the power dissipated in the resistor, and then determined how much energy had dissipated through the resistor after the switch had been opened.
hopkinmn

## Homework Statement

The problem first asks to find potential difference across the capacitor after the switch has been closed a long time. Then it asks to determine the energy stored in a capacitor when the switch has been closed for a long time. I got both these answers right, with Vcapacitor=2.22V and Energy in capacitor=4.9284*10^-6 Joules.

The last part asks to determine how much energy has dissipated through R3 after the switch has been opened.

## The Attempt at a Solution

I know that Power(P)=Current(I)*Voltage(V)=(delta charge)/(delta time)*V and that Power is the derivative of Energy. However, no charge is provided and no time is provided (we just know that it's been a long time), so I'm not sure how to go about solving this problem.

The right answer for the energy dissipated through R3 is 1.41*10^-6 Joules

#### Attachments

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There are several related expressions for the (instantaneous) power dissipated in a resistor. One involves the current through the resistor, the other the voltage across the resistor.

If you can write the equation for the voltage across the resistor w.r.t. time, then you'll be in a position to use one of those expressions for the power.

So I set up the expression P(t)=V(t)^2/R3, where V(t)=V*e^(-t/(R3*C))
When I integrate this expression, I get E(t)=-.5*V^2*C*e^(-2t/(R3*C))
Since the switch has just been opened, t=0.
I then get E(0)=-.5*V^2*C
But this just gives me the same answer for part B, where I found energy stored in the capacitor when the switch has been closed for a long time. I'm not sure what I'm doing wrong.

hopkinmn said:
So I set up the expression P(t)=V(t)^2/R3, where V(t)=V*e^(-t/(R3*C))
When I integrate this expression, I get E(t)=-.5*V^2*C*e^(-2t/(R3*C))
Since the switch has just been opened, t=0.
I then get E(0)=-.5*V^2*C
But this just gives me the same answer for part B, where I found energy stored in the capacitor when the switch has been closed for a long time. I'm not sure what I'm doing wrong.
One small problem. When the capacitor discharges it does so through both resistors R2 and R3. So the time constant must reflect that.

Ok, I understand it now, thanks so much!

## What is energy dissipation?

Energy dissipation refers to the process of converting one form of energy into another form, typically in the form of heat. This can occur when energy is transferred through a resistor, which is a device that resists the flow of electricity.

## How does a resistor dissipate energy?

A resistor dissipates energy by converting electrical energy into heat energy. This is due to the resistance within the resistor, which causes the flow of electrons to slow down and release energy in the form of heat.

## Why is energy dissipation important to understand?

Understanding energy dissipation is important because it allows us to control and manage the flow of electricity. By knowing how much energy is being dissipated through a resistor, we can ensure that the circuit is functioning properly and prevent potential hazards such as overheating.

## What factors affect the amount of energy dissipated through a resistor?

The amount of energy dissipated through a resistor is affected by its resistance, the amount of current flowing through it, and the duration of the current flow. The material and size of the resistor can also play a role in energy dissipation.

## How can energy dissipation be calculated?

Energy dissipation can be calculated using the formula P = I²R, where P is power (measured in watts), I is current (measured in amperes), and R is resistance (measured in ohms). This formula can also be used to calculate the heat generated by a resistor.

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