This discussion focuses on calculating voltage and power in RL circuits, specifically addressing the voltage across resistors and inductors. The key equations used are V=IR and P=IV, with the conclusion that the voltage drop across an inductor is zero when the current is steady. Participants clarify that since the inductor has no resistance in a steady state, the voltage across it is zero, leading to the understanding that power supplied by the current source is also zero under these conditions.
PREREQUISITESElectrical engineering students, circuit designers, and anyone studying RL circuits and their voltage and power characteristics.
You state that the current through the resistor is zero. Then what is the voltage drop across the resistor?Marcin H said:
0 volts right? Did I flip those 2 things? I thought inductors after a long time act like wires with no resistance. Is that not true?SammyS said:
Yes. zero.Marcin H said:
Ahhh. Ok that makes sense. But then how can we find how much power the current source would supply. P = IV = I^2R = V^2/R. Would we have to discharge the inductor and then use V=IR = (1A)(1000ohms)=1000V?SammyS said:
How do get 1000 ?Marcin H said:
Woops. I used the wrong equation. I meant to use P = I^2R = (1A)^2*(1000ohms) = 1000V. But is that wrong?SammyS said:
How would I get the resistance of the inductor? Are you talking about the reactance? Xl=wL? And isn't the voltage drop across the inductor 0? I'm not sure how that helps us find the power that the current source is supplying.
Yes, voltage drop across the inductor is indeed zero. Therefore, V⋅I = ?Marcin H said:
Power is 0? How can the power be 0? Does that mean the current source is just off?SammyS said:
no.Marcin H said: