Engineering Transients in 1st order RL DC circuits

[influx]

The solution states that after the switch is opened the initial current through the inductor is 20 A (in the pink box). I don't understand why its 20A. Surely the current has split at circuit junctions meaning the current arriving at the inductor is less than 20A?

Thanks

 

[.Scott]

Before the switch was opened, you had a 20A source feeding through a combination of resistors - but with a path through an inductor. Over time, the inductor will, in theory, offer no resistance, so the full current will pass through that inductor.
At the moment the switch is opened, the inductor will start to lose current, but that takes time. So at t=0, the inductor is still at 20A.

 

[influx]

Before the switch was opened, you had a 20A source feeding through a combination of resistors - but with a path through an inductor. Over time, the inductor will, in theory, offer no resistance, so the full current will pass through that inductor.
At the moment the switch is opened, the inductor will start to lose current, but that takes time. So at t=0, the inductor is still at 20A.

So you are saying that before the switch was opened, the current did not split at junctions and instead traveled straight to the inductor?

Thanks

 

[influx]

No. Before the switch is opened, the voltage across the inductor is zero, meaning no current thru any resistor. So the current flows exclusively thru the inductor.After the switch is opened, the current through the inductor cannot change instantaneously so it decays starting with 20A.

I see. One question (perhaps an obvious question), why does the voltage across the inductor equalling zero mean no current through any resistor?

Thanks

 

[rude man]

So you are saying that before the switch was opened, the current did not split at junctions and instead traveled straight to the inductor?

Thanks

Yes, because the inductor voltage before the switch is opened is zero, so no current flows thru any of the resistors.

 

[rude man]

I see. One question (perhaps an obvious question), why does the voltage across the inductor equalling zero mean no current through any resistor?

Thanks

Because I = V/R and V = 0.

 

[.Scott]

The inductor eventually becomes a short circuit.

 

[influx]

Because I = V/R and V = 0.

In the above circuit, would all the current travel through the wire (labelled with a red arrow) since it offers no resistance (theoretically)?

Thanks

 

[rude man]

In the above circuit, would all the current travel through the wire (labelled with a red arrow) since it offers no resistance (theoretically)?

Thanks

Yes.

Again, what's the voltage across the current source?
So, can there be any current in any resistor?

 

[influx]

Because I = V/R and V = 0.

The current of the inductor is given by I _inductor = V_inductor/R_inductor, so surely if the voltage across the inductor is 0, then the current through the inductor is also 0 (rather than the current in the resistors)? I am not questioning your answer, just trying to understand.

Thanks

 

[rude man]

The current of the inductor is given by I _inductor = V_inductor/R_inductor, so surely if the voltage across the inductor is 0, then the current through the inductor is also 0 (rather than the current in the resistors)? I am not questioning your answer, just trying to understand.

Thanks

Well, what about a plain old wire? It has zero volts across it by definition, but can't it carry a whopping amount of current?

If the current thru an inductor is not changing it looks exactly like a wire since V = L di/dt. Whereas for a rsistor, V = R i so that even if the current isn't changing (di/dt = 0) there must be a voltage across it. Voltage and current track each other in time perfectly in a pure resistor.