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Interview Question :: RL circuit problem


An RL circuit in which the inductance is L = 9.00 H and the resistance is R = 5.00 is connected to a 20.0 V battery at t = 0.

a) What energy is stored in the inductor when the current is 0.500 A? 1.125 J (I got this part which was relatively easy)

b) At what rate is energy being stored in the inductor when I = 1.00 A?

c) What power is being delivered to the circuit by the battery when I = 0.500 A?

Now for b and c, I was using the equation P=I^2*R=[(Blv)/R]^2 which does not really work at all for some reason on RL circuits.
Also, P=I^2R=I^2Rexp(-2Rt/L)...

Part a) I used U=1/2LI^2
I'm just very confused, can any one help please?
circuits
Answers to "RL circuit problem"
RE: RL circuit problem?

Presuming this is a complete series circuit of a battery, an inductor and a resistor.



After complete power up, this is a DC circuit, so at that point ignore the inductance part of the circuit, (L). Just use the R and voltage parts to calculate current , I=E/r and Power P=IE.



If the current delivered is less than the maximum allowed by the Resistor, the Inductor is not yet "charged" and is supplying the additional impedance on a temporary basis,

That should help with your question C. Use the measured value of I=.5a and V=20 to calculate instantaneous power.



Energy stored in a fully charged Inductor in a DC circuit will only be used when the DC changes, and will slow the decay of DC on the output if the Battery is removed. At a steady state of voltage and current, no extra energy is stored in the Inductor after steady state charge is reached. Just like a capacitor in parallel. This would be 4 amps in this case.



So how many time constants (TC) is 1amp from 0 amps on the way to 4 amps, and what was the voltage drop across 5 ohms?



Maybe this helps on why your equations don't seem to make sense.



This ignores power factor (PF) and phase issues, but those are AC parameters, not DC.
 
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