Hello, I was wondering if "pulsed" DC current behaves the same way as a steady DC current in regards to how much current is drawn in the "on or pulsed" state (in a circuit consisting of an inductor and resistor) If you have a voltage of 60VDC and a resistance of 20 ohms in a non pulsed circuit you would have current of 3 amps.

But if the current is pulsed through an inductor does it behave the same. Would it be for e.g like ( Switch ON:3 AMPS , OFF Switch OFF 0 AMPS) I know that the inductor resists a sudden change in current so by obvious reasons it's not that straight forward but in general on average would it be around the same?

 

Depends on how much inductance there is and how short the pulse is.

If the circuit is reasonably modelled by a resistor in series with an inductor, there is a time constant given by L/R. If that time constant is short compared to the width of the pulse (say 1/10 or less), then the response will be pretty close to "instant" on and off. But if the L/R time constant is not short compared to the pulse width, then it will have a major effect on the response. In the case where it is much larger than the pulse width, you would see only a small fraction of the DC current.

 

An inductor and a resistor have two different analyses that can be performed. These are the transient response and the steady state response. In the steady state response Ohm's Law applies, just as in a DC circuit. In the transient response there is a 1st order differential equation that governs the response. That differential equation for the response to a unit step function is:

\( L\cfrac{di}{dt}\;+\;iR\;=\;u(t) \)

As you can see, when steady state is achieved, the derivative term will vanish, and Ohm's law is what is left.

 

But if the current is pulsed through an inductor does it behave the same?

No.
Assuming the R (20 Ohms) and L (1 Henry) are in series, here's what you'd get if a 60V source were pulsed with on/off times of 10mS each :-

Note the initial current rise and that the long term average current is about 1.5A, because the pulse duty cycle is 50%.

 

No.
Assuming the R (20 Ohms) and L (1 Henry) are in series, here's what you'd get if a 60V source were pulsed with on/off times of 10mS each :-
View attachment 288521
Note the initial current rise and that the long term average current is about 1.5A, because the pulse duty cycle is 50%.

And each of those (apparently) linear segments of the current waveform follow the differential equation given in post #3 with different initial conditions. The forcing function changes from 60u(t) to -60u(t) every 10 milliseconds. The final value of one solution becomes the initial condition for the next solution.

For more details on the RL circuit, read the following article:
https://en.wikipedia.org/wiki/RL_circuit

The L/R time constant for this circuit is 1H/20Ω = 0.05 seconds

This period of the input is "short" with respect to the time constant, so you only get a small change in the current during the "on" time of the input. That is why you see the behavior shown by the simulation.

 

And each of those (apparently) linear segments of the current waveform follow the differential equation given in post #3 with different initial conditions. The forcing function changes from 60u(t) to -60u(t) every 10 milliseconds. the final value of one solution becomes the initial condition for the next solution.

For more details on the RL circuit, read the following article:
https://en.wikipedia.org/wiki/RL_circuit

The L/R time constant for this circuit is 1H/20Ω = 0.05 seconds

This period of the input is "short" with respect to the time constant, so you only get a small change in the current during the "on" time of the input. That is why you see the behavior shown by the simulation.

Thank you so much!, that makes sense.