Power in the components of an RC circuit as a Function of time

[physninj]

Homework Statement

A simple RC circuit with an EMF source ε, Resistor R and capacitor C are in series. When the switch is closed find the rate of energy stored in the capacitor, the power dissipated in the resistor, and the power supplied by the emf as functions of time.

Homework Equations

For the resistor

P=i²R

i=(ε/R)e^-t/RCFor the battery

P=iε

i=(ε/R)e^-t/RC

For the capacitor

q=CE(1-e^-t/RC)

U=q²/(2C)

The Attempt at a Solution

I feel decent about the expressions I got for the resistor and the battery.

P_res=[(ε/R)e^-t/RC]²R

P_bat=[(ε/R)e^-t/RC]εThe problem I'm having is getting the energy rate in the capacitor. I tried to just use the energy equation from above and divide by time but I don't think it's right, It's undefined at t=0, and I need to evaluate these equations at that point in later steps.

I think I may need to plug it into the energy and then take the time derivative. Its pretty messy though, is there a better way?

[CE(1-e^-t/RC)]²/(2Ct)

Thanks for any input

 

[SammyS]

Homework Statement

A simple RC circuit with an EMF source ε, Resistor R and capacitor C are in series. When the switch is closed find the rate of energy stored in the capacitor, the power dissipated in the resistor, and the power supplied by the emf as functions of time.

Homework Equations

For the resistor

P=i²R

i=(ε/R)e^-t/RC


For the battery

P=iε

i=(ε/R)e^-t/RC

For the capacitor

q=CE(1-e^-t/RC)

U=q²/(2C)

The Attempt at a Solution

I feel decent about the expressions I got for the resistor and the battery.

P_res=[(ε/R)e^-t/RC]²R

P_bat=[(ε/R)e^-t/RC]ε


The problem I'm having is getting the energy rate in the capacitor. I tried to just use the energy equation from above and divide by time but I don't think it's right, It's undefined at t=0, and I need to evaluate these equations at that point in later steps.

I think I may need to plug it into the energy and then take the time derivative. Its pretty messy though, is there a better way?

[CE(1-e^-t/RC)]²/(2Ct)

Thanks for any input

There are several ways to find the rate of energy stored in the capacitor.

You can use conservation of energy. You know the power supplied by the battery (rate of work done by the battery). You know the power dissipated in the resistor. From those it's fairly direct to account for the rest of the energy.

Alternatively, you can find the voltage drop across the resistor as a function of time. From that you can find the voltage drop across the capacitor as a function of time. Use that with U = (1/2)V²C.

You can integrate the current to find the charge on the capacitor.

...

 

[physninj]

There are several ways to find the rate of energy stored in the capacitor.

You can use conservation of energy. You know the power supplied by the battery (rate of work done by the battery). You know the power dissipated in the resistor. From those it's fairly direct to account for the rest of the energy.

Alternatively, you can find the voltage drop across the resistor as a function of time. From that you can find the voltage drop across the capacitor as a function of time. Use that with U = (1/2)V²C.

You can integrate the current to find the charge on the capacitor.

...

I see, because what's not dissipated by the resistor is stored in the capacitor. I think I will take that route.

So its just

P_emf-P_resistor=P_capacitor

My other equations look alright I'm assuming?Thanks a bunch.

 

[SammyS]

I see, because what's not dissipated by the resistor is stored in the capacitor. I think I will take that route.

So its just

P_emf-P_resistor=P_capacitor

My other equations look alright I'm assuming?


Thanks a bunch.

Yes.

They look right.

 

[Nickie]

or advice.I would approach this problem by first understanding the physical principles involved in an RC circuit. The resistor and capacitor are connected in series, which means that the current through each component is the same. The capacitor stores energy in the form of electric charge, while the resistor dissipates energy in the form of heat. The EMF source provides the energy to keep the circuit running.

To find the rate of energy stored in the capacitor, we can use the equation for the energy stored in a capacitor: U = q^2/(2C). We can differentiate this with respect to time to get the rate of energy stored: dU/dt = (2q/C)(dq/dt). We can substitute in the expression for the charge on the capacitor (q = CE(1-e^-t/RC)) and its derivative (dq/dt = -CE/R * e^-t/RC) to get dU/dt = -E^2 * e^-2t/RC. This expression gives us the rate of energy stored in the capacitor as a function of time.

To find the power dissipated in the resistor, we can use the equation P = i^2R. Since the current through the resistor is the same as the current through the capacitor, we can use the expression for the current through the capacitor (i = E/R * e^-t/RC) and substitute it into the power equation to get P = E^2 * e^-2t/RC. This gives us the power dissipated in the resistor as a function of time.

Finally, to find the power supplied by the EMF source, we can use the equation P = iE. Again, we can use the expression for the current through the capacitor and substitute it into the power equation to get P = E^2 * e^-2t/RC. This gives us the power supplied by the EMF source as a function of time.

Overall, we can see that the power in the components of an RC circuit changes over time due to the behavior of the capacitor. Initially, when the switch is closed, the capacitor is uncharged and there is a large amount of energy being supplied by the EMF source. As time goes on, the capacitor charges up and the power supplied by the EMF source decreases while the power dissipated in the resistor also decreases. Eventually, the capacitor reaches its maximum charge and the power supplied by the EMF source