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We know that the inductor creates an emf trying to oppose the change in ΔΦ. For an AC LR circuit I found the following diagram.enter image description here

My question is why doesn't the current reach its peak a little earlier than π/2? When the voltage is at its peak I suppose the emf is smaller than the source voltage so the current keeps increasing, trying to flow through the inductor. The emf is big at this point based on what I know for the DC RL circuit and decreases through time. We can tell from the diagram that the current increases until π/2 when the inductor comes into play to oppose its decrease this time.

However, I can't understand why this is not happening earlier. I get the increase in current at first when voltage is at maximum value but when it gets really small the current has already reached the maximum value for that particular voltage. So won't the emf change polarity at that point?

In case I wasn't clear I give you this example. Say we have a DC LR circuit. Time has passed and the current has reached its maximum value given by Ohm's law. Now I disconnect the source and connect another source with a lower voltage(this happens instantly). The current will slowly decrease because the emf tries to keep it constant. So the decrease starts now and that is because the voltage is lower.

Isn't there a time before π/2 in the AC circuit when voltage is lower than the voltage given by Ohm's law for the current at that moment?

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  • \$\begingroup\$ this is how current energy is stored in an inductor with a 90 deg lag in current. thus the real power V(t)*I(t) over 1 cycle= 0 and this is stored reactive energy \$\endgroup\$ – Sunnyskyguy EE75 Jan 28 '17 at 23:59
  • \$\begingroup\$ Hmm. Are you trying for a deep understanding that "sings in your mind" so to speak? Or would it be acceptable to just tell you to look at the curves and notice that the peak magnitude of one takes place when the other one is at its maximum rate of change? \$\endgroup\$ – jonk Jan 29 '17 at 0:15
  • \$\begingroup\$ Deep understanding. I can read the diagram. \$\endgroup\$ – John Katsantas Jan 29 '17 at 2:58
  • \$\begingroup\$ The voltage across the inductor is a result of the rate of change of current through the inductor. Not the other way round. Ohms law is not a good way to analyse it. \$ V = -L\frac{di}{dt} \$ \$\endgroup\$ – crowie Jan 29 '17 at 4:49
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    \$\begingroup\$ You might want to look at my answers to these two questions. The first shows plots of V and I, but starting from zero, so you see the difference between transient/DC behaviour, and long term AC electronics.stackexchange.com/questions/281476/…. The second talks about capacitors as the dual of inductors, to aid understanding, if you grok the first, this might help you grok the second. electronics.stackexchange.com/questions/282053/… \$\endgroup\$ – Neil_UK Jan 29 '17 at 6:21
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The diagram is for the pure inductance only. In other words, the voltage shown there is the voltage across the inductor only, and the current is only the inductor current. These values are always 90° out of phase.

When considering the series or parallel combination of an inductor with a resistor, it gets a little more complicated. In a series circuit, the current through both components is identical, but the voltages have different magnitudes and phases, and their sum has a phase angle of somewhere between 0° and 90° relative to the current.

Similarly, in a parallel circuit, the voltage across both components is identical, but the currents have different magnitudes and phases, and their sum has a phase angle of somewhere between 0° and 90° relative to the voltage.

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As already commented, your image is for pure inductance, no resistors included. But that does not change the explanation.

The inductor has slow response. When a voltage is applied over it, the current starts to rise.

It rises as long as the voltage has the original polarity. The voltage can get lower, but as long as it has the original polarity, the current in the inductor gets higher.

The tide turns when the polarity of the voltage changes. Then inductor's current starts to decrease. It can still have its original direction quite long time.

In mathematics we say that the growth rate (amperes/second) of inductor's current is = the voltage over the inductor divided by the inductance (Henrys)

In theory one volt connected to one Henry inductor should result infinitely growing current that has growth rate = 1 amperes /second. In practice all electric parts have some resistance. Thus the growth rate of the current tails off and the current never goes over the Ohmic limit - that's equal to voltage/total resistance.

probably you have seen this:

How does the inductor ''really'' induce voltage?

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