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A coil with zero resistance has its ends labeled a and b. The potential at a is higher than at b. Which of the following could be consistent with this situation?

Mark all that apply

a. The current is constant and is directed from a to b.

b. The current is constant and is directed from b to a.

c. The current is decreasing and is directed from b to a.

d. The current is increasing and is directed from a to b.

e. The current is increasing and is directed from b to a.

f. The current is decreasing and is directed from a to b.

The correct answers are C and F, I don't understand why, I know current goes from higher to lower potential and that an induced emf is opposite the direction of the battery. Also, how can there be a potential difference with zero resistance ?

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closed as off-topic by John Rennie, Jon Custer, Emilio Pisanty, rob Aug 10 '18 at 19:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Thank you to all the people who helped me with this problem! $\endgroup$ – Francesco Benvenuto Aug 14 '18 at 17:42
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First, let's clear up a misconception:

I know current goes from higher to lower potential

That's not necessarily true. It's always true for resistors but not always true for, e.g., an inductor which is discharging (supplying energy to the attached circuit).

Also, how can there be a potential difference with zero resistance?

Since there can be no electric field within an ideal conductor, the mobile charge distribution within the conductor will change depending on any emf due to a changing magnetic flux threading the coil. The conservative electric field of the mobile charge distribution will be just what is needed to cancel, within the conductor, the non-conservative field electric field that gives rise to the emf. This conservative electric field gives rise to a potential difference across the inductor.

The correct answers are C and F, I don't understand why

I don't either. Let me explain...

Since there is a voltage across the inductor, the current through must be changing and so the answer cannot be (a) or (b).

Assuming the inductor voltage is defined as $v_L = v_{ab}$, the voltage across the inductor is positive if the potential at $a$ is higher than at $b$.

By the passive sign convention, a positive inductor current is directed from $a$ to $b$ through the inductor (note that by current, I mean electric current and not electron current).

Since the voltage across the inductor is positive by stipulation, the inductor current is increasing (becoming more positive).

So, if the current is initially positive (directed from $a$ to $b$), the current is increasing (becoming more positive) which is answer (d).

But if the current is initially negative (directed from $b$ to $a$), the current is still increasing (becoming more positive) which is answer (e).

I suppose one might claim that the negative current is becoming less negative and, in this sense, decreasing which would be answer (c).

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If you have an ideal inductor it cannot have a potential difference across it if the current is not changing so options (a) and (b) are not correct.

To decide on the other options I think it helps to add extra components to the inductor and complete an electrical circuit.

enter image description here

Consider circuit 1 where the initial current through the inductor is zero and then the switch is closed.
An increasing current will flow in the direction shown in the diagram and the potential of node a is certainly higher than that of node b because of the orientation of the voltage supply.
So option d is correct whilst option e (reverse the polarity of the voltage supply) is not.

Now consider circuit 2 with the current flowing as shown in the diagram.
The current will be decreasing and the potential of node a will be greater than that of node b as current always flows through a resistor from the higher potential to the lower potential.
So option c is correct whilst option f (reverse the direction of the current) is not.

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  • $\begingroup$ I think you mean if you have an ideal inductor it cannot have a potential difference across it if the current is not changing (e.g., dc ). $\endgroup$ – Bob D Aug 5 '18 at 12:40
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    $\begingroup$ @AlfredCentauri I have added the words referring to an unchanging current as suggested by BobD. $\endgroup$ – Farcher Aug 5 '18 at 13:32
  • $\begingroup$ @BobD Many thanks for your suggestion which I have incorporated in my answer. $\endgroup$ – Farcher Aug 5 '18 at 13:33
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Consider a sinusoidal current involving an ac voltage in series with a resistor and inductor. See the diagrams below showing voltage across and current through the inductor for each quarter cycle. Based on the diagrams, choices (d) and (e) are correct.

Answers (a) and (b) are incorrect since an ideal inductor looks like a short circuit to constant current. enter image description here

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  • $\begingroup$ "Regardless of whether the current is increasing (first quarter cycle) or decreasing (second quarter cycle) the polarity is + at a and – at b." - I don't believe this is correct. The sign of the voltage across an inductor is positive if the current through is increasing and negative if the current through is decreasing. $\endgroup$ – Alfred Centauri Aug 6 '18 at 0:44
  • $\begingroup$ You are right. Got myself confused without using diagrams. See my edited answer. $\endgroup$ – Bob D Aug 6 '18 at 2:42

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