DC Circuit and an Inductor - Q on potentials

[milky9311]

Homework Statement

While the voltage induced between the points P and Q is changing, which is the high potential end of the inductor? Explain your answer.

P.S. sorry for the fail Paint job lol

Homework Equations

The Attempt at a Solution

I know that in a DC circuit current flows from the higher potential (left on diagram) anti clockwise to the other side of the source.

I also know that inductors induce a back emf that resists the supply voltage, so i thought that Q would be higher potential as to oppose the supply voltage but the answer says P! I don't know why!

Thanks for your help

 

[NascentOxygen]

That third element is a switch, is it? When the switch is opened, Q will go very positive (with respect to all other points in the circuit).

 

[milky9311]

That third element is a switch, is it? When the switch is opened, Q will go very positive (with respect to all other points in the circuit).

i think the question asks when the switch was closed, i just dfon't know why P would be the higher potential when a back emf is induced by the inductor

 

[NascentOxygen]

When the switch is closed, the potential across the inductor is a constant 12 volts. So voltage is not changing. Induced emf while (ideal) battery is supplying current is not changing, either--induced emf exactly equals a constant 12 volts at all times while the switch is closed.

Question refers to a time when induced emf is changing. The only time this happens is when the switch is opened and interrupts the current.

 

[gneill]

i think the question asks when the switch was closed, i just dfon't know why P would be the higher potential when a back emf is induced by the inductor

Consider that in the instant after the switch is closed, the inductor will not have allowed any current to start to flow (inductors are stubborn that way ). The back-emf that the inductor produces must oppose the voltage source trying to drive current through it. So P will display a voltage equal to that of the driving potential (the battery).

Another way to look at it is that at the instant that the switch closes, the inductor "looks like" an open circuit. What then would be the voltage at P? At Q?

So while the switch is close the voltage across the inductor is fixed at the battery voltage.

When the switch opens, interesting things happen The inductor wants to keep the current flowing that it had just before the switch was opened. To accomplish this it will generate any voltage required in that first instant (usually this leads to arcing between the switch contacts). The inductor draws on its stored energy (in the form of its magnetic field) to do this.

In order to keep try to the current going in the same direction, what should be the polarity of the voltage across the inductor?

 

[NascentOxygen]

when the switch was closed, i just dfon't know why P would be the higher potential when a back emf is induced by the inductor

When and while the switch is closed, P is +12volts wrt point Q. The battery voltage sets this, regardless of the element being a resistor, capacitor, or inductor. (I'm assuming ideal elements.)