Understanding a Simple RL Circuit | GRE Physics Problem Explanation

[WarPhalange]

Homework Statement

The problem is from an old GRE that I can't wrap my head around.

http://grephysics.net/ans/8677/94

Homework Equations

V = -L(dI/dt)
I = V/R

The Attempt at a Solution

Okay, so I have the solution, I just don't understand their reasoning. A change in current through an inductor makes it produce a voltage. I thought it was in the opposite direction of where current was coming from, though, meaning that it would make the voltage across A smaller at first and eventually let up and let the voltage go up to its asymptotic value.

Then when you closed the switch, I understand that the only way for the built up charge to dissipate is to make it go in a loop, through A, L, R2, and the Diode. I don't understand why it doesn't all just go directly from L to Ground?

 

[MATLABdude]

Homework Statement

The problem is from an old GRE that I can't wrap my head around.

http://grephysics.net/ans/8677/94

Homework Equations

V = -L(dI/dt)
I = V/R

The Attempt at a Solution

Okay, so I have the solution, I just don't understand their reasoning. A change in current through an inductor makes it produce a voltage. I thought it was in the opposite direction of where current was coming from, though, meaning that it would make the voltage across A smaller at first and eventually let up and let the voltage go up to its asymptotic value.

Then when you closed the switch, I understand that the only way for the built up charge to dissipate is to make it go in a loop, through A, L, R2, and the Diode. I don't understand why it doesn't all just go directly from L to Ground?

The polarity of the induced voltage is such that it maintains the current through the inductor. In this case, the initial current (at t=0) is 0. The only way to do this is for the induced voltage (EMF) to be equal to the battery voltage, but with the "top" more positive than the "bottom" of the inductor.

Also, there is no magic "ground" that all your charges return to. Ground is just a reference, and usually a wire, or the point at which you put the negative terminal of your DMM. You can make this approximation because voltage sources "pump" the same amount of current out of ground as goes into it. In short, a wire.

Hope this helps!

 

[WarPhalange]

The polarity of the induced voltage is such that it maintains the current through the inductor. In this case, the initial current (at t=0) is 0. The only way to do this is for the induced voltage (EMF) to be equal to the battery voltage, but with the "top" more positive than the "bottom" of the inductor.

Yeah, that explanation helps a bit... but I still don't get why voltage is highest at point A when you first close the switch. If the EMF is pushing against the battery voltage, wouldn't you get 0V at point A?

I understand that after you open the switch, the voltage has to dissipate in a loop, so it is negative at point A, positive going through the diode, and then it goes to 0 eventually.

Also, there is no magic "ground" that all your charges return to. Ground is just a reference, and usually a wire, or the point at which you put the negative terminal of your DMM. You can make this approximation because voltage sources "pump" the same amount of current out of ground as goes into it. In short, a wire.

Hope this helps!

I thought if something was connected to ground, it automatically had it's voltage be at the ground voltage no matter what. Would taking out the connection to ground in this circuit change anything at all?

 

[MATLABdude]

Yeah, that explanation helps a bit... but I still don't get why voltage is highest at point A when you first close the switch. If the EMF is pushing against the battery voltage, wouldn't you get 0V at point A?

I understand that after you open the switch, the voltage has to dissipate in a loop, so it is negative at point A, positive going through the diode, and then it goes to 0 eventually.

(NOTE: In this post, EMF stands for the value of the voltage source)

No, consider the potential difference between point A and the positive terminal of the battery. If A were -EMF (relative to ground), this potential difference would be -2EMF, and a current would flow through the resistor. If A were +EMF (relative to ground), then this potential difference is 0, and no current would flow through the resistor (as is the case).

I thought if something was connected to ground, it automatically had it's voltage be at the ground voltage no matter what. Would taking out the connection to ground in this circuit change anything at all?

Sort of... "Ground" is just a reference point (so that a potential of, say, 10V at some point in a circuit means that this point has 10V of potential relative to your ground point). You could very well set your ground point to be at the top of the battery, and the bottom of the battery would be -EMF. This is tantamount to switching your voltmeter probes around (and the resulting graph would be the negative of the one given).

Taking out the "connection" to ground would mean that you have nothing to measure potential relative to. This might be EE-centric, but it's like trying to take a voltage measurement using only one probe of your voltmeter--something impossible to do (unless your other probe is somehow connected to your circuit).

 

[WarPhalange]

(NOTE: In this post, EMF stands for the value of the voltage source)

No, consider the potential difference between point A and the positive terminal of the battery. If A were -EMF (relative to ground), this potential difference would be -2EMF, and a current would flow through the resistor. If A were +EMF (relative to ground), then this potential difference is 0, and no current would flow through the resistor (as is the case).

God that's confusing... so it's +EMF until it hits something with resistance, where it goes to 0 in order for there not to be any current in the circuit?

So if there were to R1 resistors with some wire in between instead of 1 R1 resistor like it is now, the EMF would be 0 in between them when the switch was first closed?

Taking out the "connection" to ground would mean that you have nothing to measure potential relative to. This might be EE-centric, but it's like trying to take a voltage measurement using only one probe of your voltmeter--something impossible to do (unless your other probe is somehow connected to your circuit).

You have a plus and a minus terminal on the batteries with a voltage difference between them. Wouldn't that be enough?

 

[MATLABdude]

God that's confusing... so it's +EMF until it hits something with resistance, where it goes to 0 in order for there not to be any current in the circuit?

So if there were to R1 resistors with some wire in between instead of 1 R1 resistor like it is now, the EMF would be 0 in between them when the switch was first closed?

No, it's +EMF all the way from the +ve voltage terminal through until point A (both measured relative to ground). So there is 0 potential difference between one end of the resistor and the other (or the ends of however many resistors you have in a row), and hence no current flow. Whether or not current flows across a resistor depends on the voltage difference between the ends of said resistor. This relates to your point below:

You have a plus and a minus terminal on the batteries with a voltage difference between them. Wouldn't that be enough?

No, you'd still have to measure relative to something. That something (in most introductory circuit analysis / electromagnetics classes) is just (sometimes implicitly) taken as the negative terminal of the voltage supply. But usually this is made explicit with the ground symbol (unless someone dropped the ball somewhere).

 

[WarPhalange]

Cool, makes perfect sense now. Thanks a lot MATLABdude.