Energy dissipated by a resistor in an RC circuit

In summary: I(rms)=V(rms)/R=sqrt(P/R). so that V(rms)=sqrt(1/C integral(P dt)). In summary, we have a series circuit with a capacitor, resistor, and DC voltage source. At t=0, the switch is closed and the capacitor begins to charge. Eventually, the capacitor becomes fully charged and the current decreases to 0. To find the energy dissipated by the resistor at t=infinity, we can use the equation PR=i^2 R and integrate from t=0 to t=infinity. Alternatively, we can solve for the voltage drop across the resistor as a function of time and integrate from t=0 to t=infinity. Both methods give the same answer of CV^
  • #1
ehilge
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Homework Statement


Given a series circuit with a capacitor of capacitance C, a resistor of resistance R, a DC voltage source with a voltage V, and an open switch (by open I mean not connected so no current flows through). At t=0 the switch is closed. How much energy is dissapated by the resistor at t= infinity

Homework Equations


C=Q/V
PR=iV
V=iR
i=dq/dt

The Attempt at a Solution


First off my conceptual understanding of the problem is such: Initially, the voltage source provides a voltage that charges up the capacitor. Eventually, the capacitor charges all of the way up and the current goes to 0. So what I need to do is integrate the power that the resistor dissipated from t=0 to t=infinity
Using PR=iV and subbing in i=dq/dt.
The voltage can factor our of the integral since it is constant, and the dt's cancel
I end up with V times the integral of dq which equals VQ
Since Q=CV the answer I have is CV2

Now I have no idea if this is correct since it is a test review question so I was wondering if this looks correct to all of you. If not, as always, any input would be greatly appreciated.
Thanks!
 
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  • #2
You could do it a second way to check your answer. Solve for v(t) of the capacitor, and then write the expression for the voltage drop across the resistor as a function of time. Then integrate that voltage difference across the resistor from 0 to infinity. Do you get the same answer?
 
  • #3
The voltage across the resistor is not constant! V(resistor)=iR. This gives P=i^2 R.
 

FAQ: Energy dissipated by a resistor in an RC circuit

How is energy dissipated by a resistor in an RC circuit?

In an RC circuit, energy is dissipated by the resistor through the process of Joule heating. As the current flows through the resistor, it encounters resistance, which causes some of the electrical energy to be converted into heat energy. This heat energy is then dissipated into the surrounding environment.

What factors affect the amount of energy dissipated by a resistor in an RC circuit?

The amount of energy dissipated by a resistor in an RC circuit is influenced by several factors, including the resistance of the resistor, the voltage applied to the circuit, and the duration of the current flow. A higher resistance or voltage will result in more energy being dissipated, while a shorter duration of current flow will result in less energy being dissipated.

How is the energy dissipated by a resistor in an RC circuit calculated?

The energy dissipated by a resistor in an RC circuit can be calculated using the formula E = I2Rt, where E is the energy dissipated in joules, I is the current in amperes, R is the resistance in ohms, and t is the time in seconds. This formula assumes that the voltage is constant over time.

Can the energy dissipated by a resistor in an RC circuit be recovered?

No, the energy dissipated by a resistor in an RC circuit cannot be recovered. Once the electrical energy is converted into heat energy and dissipated into the environment, it cannot be converted back into electrical energy. This is due to the law of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.

How can the energy dissipated by a resistor in an RC circuit be reduced?

The energy dissipated by a resistor in an RC circuit can be reduced by decreasing the resistance of the resistor or reducing the voltage applied to the circuit. Additionally, using a resistor with a higher power rating can help to dissipate more energy without overheating. It is also important to limit the duration of the current flow to minimize the amount of energy dissipated by the resistor.

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