Energy dissipated by a resistor in an RC circuit

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SUMMARY

The energy dissipated by a resistor in an RC circuit, when a switch is closed at time t=0, is calculated using the formula CV², where C is the capacitance and V is the voltage of the DC source. Initially, the capacitor charges until the current reaches zero, at which point the resistor dissipates energy. The integration of power over time confirms this result, as the voltage across the resistor is a function of the current, which changes over time. Alternative methods, such as solving for v(t) of the capacitor and integrating the voltage drop across the resistor, yield the same conclusion.

PREREQUISITES
  • Understanding of RC circuits and their components (capacitor, resistor, voltage source)
  • Knowledge of basic electrical equations (Ohm's Law, power equations)
  • Familiarity with calculus, particularly integration
  • Concept of energy conservation in electrical circuits
NEXT STEPS
  • Study the derivation of the energy stored in a capacitor (Energy = 1/2 CV²)
  • Learn about the time constant in RC circuits and its effect on charging/discharging
  • Explore the relationship between current and voltage in resistors over time
  • Investigate alternative methods for analyzing transient responses in circuits
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone preparing for exams involving circuit analysis and energy dissipation in electrical components.

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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!
 
Last edited:
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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?
 
The voltage across the resistor is not constant! V(resistor)=iR. This gives P=i^2 R.
 

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