If an inductor is connected to a power for example via a switch it self-induces a voltage, limiting the current flow through it according to Lenz's Law.
I imagine this as connecting another voltage source to the circuit, but with opposite polarity to the supply voltage reducing the overall voltage in the circuit and thus the current through the inductor.
If you connect another component in parallel to the inductor (for example a lamp or a resistor) there should be a drop in voltage across that branch as well, because it is parallel to the inductor (so our model second voltage source). This should also mean a reduction in current through the branch in parallel, but (at least as far as we got in physics lessons) this doesn't happen.
The voltage drop and thus the current drop is somehow limited to the inductor itself.
Why is this though? Doesn't this hurt very basic principles of electronics or is the model with the second voltage source flawed or are there effects, that are just too small or too complicated to be considered in school physics?
No. The voltage is set by the voltage source. The components draw current, according to the applied voltage, and their particular characteristics.
The above assumes an ideal voltage source, with zero internal resistance. A real voltage source will behave differently.
Because an inductor creates a magnetic field, the current limit is only a change in current that is limited. The RESISTANCE of the inductor limits the steady current.
Why self-induced voltage only limited to the inductor
Because it's the only passive device that can generate a self-induced voltage
The voltage across the inductor and the other component is as supplied by the voltage source. For the inductor, if the current is changing then so will be the magnet field, which will produce a voltage such as to oppose the change in current. Or to put it another way an inductor will act to try to keep the current through it constant. All this is within the inductor, it has no effect on other components in parallel with it.
Imagine a variable voltage source in series with the inductor, actually built into the insides of the inductor. That voltage will add to the externally applied voltage, changing the voltage the inductor sees, and thus the current through it. The other components don't see this voltage source so are unaffected.
Excellent question by the way, well done for asking it.
Thank you very much for your answer! But why if there is (as a model) a second voltage source, is it contained within the inductor. If we look at the moment the inductor is supplied a voltage, the self-induced voltage equals the supply voltage, but negative. If there was actually a second voltage source, shouldn't that compensate the supply voltage across the whole circuit?
Why isn't that the case? We did an experiment with a lamp parallel to the inductor through which the current stayed constant the entire time, while I would expect it to rise together with the current through the inductor, because I would expect the voltage across the entire circuit to first rise according to the current change in the inductor.
I would expect the voltage across the entire circuit to first rise according to the current change in the inductor.
It seems that you have not yet understood or taken the rules of ideal circuits to heart.
See reply #2 and review your textbook.
We did an experiment with a lamp parallel to the inductor through which the current stayed constant the entire time, while I would expect it to rise together with the current through the inductor
The result clearly demonstrates that your expectation is wrong. Good experiment!
There is a voltage source 'inside' the inductor, it is there whenever the magnetic field changes. The 'inside' voltage source is as a result of the magnetic field interacting with the inductor. It's the same as you would get if you had an externally produced magnetic field moving through the inductor, the inductor does not care where the field comes from, all it sees is a changing magnetic field. Changing field -> induced voltage. The induced voltage will be what it needs to be to keep the current from changing. To be clear: This only happens when the current and thus the field is changing. So, when you first apply voltage some current begins to flow, there is a weak but increasing field (so changing), which induces a voltage of opposite polarity to the applied voltage so reduces the effective voltage across the inductor's resistance, limiting the current. However, this only works while the current, and thus the field, is changing. Changing in this case means increasing, so as the current / field increases the effect gets less until there is no significant induced voltage and the current stabilises according to Ohms law and the resistance of the coil.
I don't know how to explain better.
Please draw this and indicate what happened and what you thought would happen.
I appreciate your help! However I think you misunderstood my question and the intent of this question. I am fully aware that my expectation is wrong. I am trying to understand why. And I am (at least in my probably somewhat distorted self-awareness) aware of the basic laws of circuitry. These however tell me, that in a parallel circuit the voltages across the branches are equal. There is a reduced current through the inductor, indicating a voltage drop across the branch with the inductor. This voltage drop should be visible across all branches in parallel, but it isn't. That's why I am so confused!
Yes, from V0 (set by the voltage source) to 0 (GND), always. So the current through the lamp is constant.
I realise that you were replying to @jremington not me, but read again my description of a second voltage source inside the inductor.
Thank you very much for sticking around! 
This is the circuit I was referring to (ignore the 60 ohms)
The two branches (I will call them B1 for the inductor branch with I1 and B2 for the other one) are parallel to each other. I close S.
I expect the following: I1 starts at 0A and slowly rises. This must be caused by a reduced voltage across B1. This means because B1 and B2 are parallel, this voltage drop must also apply to B2 (If there is a reduced flow of current through B1, there must be a reduced voltage across it), meaning I2 should behave identically to I1 (if R1=R2).
What happens: I2 just stays constant, the current change is limited to B1, so only I1 changes.
How can have two parallel branches two different voltages across them?
Last reply before I go to bed.
The voltage across the 2 branches comes from the supply on the left. That voltage is fixed by the supply and does not change. Both branches therefore get the same voltage.
The current through R2 is, as I think you know, the supply voltage / R2.
The current through L is (the supply voltage + the induced voltage in the inductor*) / R1
Now go and think about this and the other answers you have.
Do you know how a transformer works? If not look it up.
Do you know Kirchhoff's circuit laws - Wikipedia ? If not then learn them.
*Remember this voltage will have the opposite polarity to the supply voltage.
They don't. The voltage drop across them is always V0, the supply voltage.
Keep at it, and eventually the concepts will sink in.
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