Register to reply 
Power in a driven RLC Circuit 
Share this thread: 
#1
Mar612, 06:20 PM

P: 15

1. The problem statement, all variables and given/known data
For a driven RLC circuit, compare the power delivered by the source to the power dissipated as heat in the resistor. 2. Relevant equations P[itex]_{avg}[/itex] = I[itex]_{rms}[/itex]*V[itex]_{rms}[/itex]*cos([itex]\phi[/itex]) 3. The attempt at a solution My thinking was that the power dissipated in the resistor would be less than the power delivered by the source. I thought so because in addition to the energy radiated as heat at the resistor, energy is either being stored or being released back into the circuit by the capacitor and inductor. However, I was told that the power at the source equals the power dissipated as heat in the resistor. If all of the source energy is accounted for by the capacitor, what happened to the energy in the inductor and capacitor? 


#2
Mar612, 06:51 PM

Mentor
P: 11,678




Register to reply 
Related Discussions  
Interruptdriven circuit  Electrical Engineering  1  
Power of a driven oscillating spring  Introductory Physics Homework  1  
Thorium Accelerator Driven Nuclear Power  Why not ?  Nuclear Engineering  37  
Driven RL circuit, why at infinity is this the current 2, and yet at 0+, its 0?  Introductory Physics Homework  4  
Efficiency of a power driven car  Introductory Physics Homework  6 