Inductance in AC circuit Why does an inductor oppose a change in current and not voltage? Is it a rule or is it because of lenz's law? Because intuitively the back e.m.f which opposes a change in current should also oppose a change in voltage? Pls help
 A: In a simple circuit where there is a changing voltage source with an inductance and a resistance in series, the same current flows through every device, so it is clear to say that the back emf opposes the current. On the other hand voltage levels can change in different ways at different points in different parts of the circuit, so it would not be clear to say that the back emf opposes a change in voltage.
When a power switch is turned off sparks are sometimes seen in the switch. This is caused by the sudden change in current, which produces a large emf opposing the change in current. This short-lived emf can be much larger than the supply voltage, which is why it sparks. Talking about the emf opposing the change of voltage in such a case would be especially difficult to understand.
A: 
Why does an inductor oppose a change in current and not voltage?

Because induced electric field (which causes induced EMF) is associated with changing current (accelerated motion of electric charges), not changing potential difference (voltage). This is one of EM laws. The charges have to accelerate for induced electric field to be present, and the induced electric field is then proportional to magnitude of this acceleration.

Because intuitively the back e.m.f which opposes a change in current should also oppose a change in voltage?

No. What is intuitive or not is subjective. We can't rely on intuition, we need to seek laws experimentally and mathematically.
Experimentally, the "back-emf" is well explained by just the induced EMF due to induced electric field, proportional to magnitude of acceleration of charges. We know (EM laws) that induced EMF in an inductor is given by
$$
\mathscr{E}_i = -L\frac{dI}{dt}.
$$
If the inductor is perfect (zero resistance and capacitance), voltage (drop of potential) on its terminals is
$$
V = -\mathscr{E}_i = L\frac{dI}{dt}.\tag{*}
$$
What you're suggesting is that there may be another EMF $\mathscr{E}_T$ proportional to rate of change of voltage and acting against it:
$$
\mathscr{E}_{T} = -A\frac{dV}{dt}.
$$
where $A$ is some constant. This may be consistent with experimental knowledge if $A$ is very small, so that (*) remains accurate.
It would mean that
$$
\mathscr{E}_{T} = -AL\frac{d^2I}{dt^2}\tag{**}
$$
i.e. this hypothetical EMF would be proportional to second derivative of current, or in other words, proportional to first derivative of acceleration of charges. Ordinary (non-radiating) circuits are well explained without it. So there is no need to introduce it and complicate the theory.
If we wanted to model a radiating circuit, there would be new EMF accounting for damping of current due to radiation, which would look similar to (**). But the sign would be the opposite; radiation damping is proportional to rate of change of acceleration, so it would support the change in voltage, not oppose it.
A: 
Why does an inductor oppose a change in current and not voltage?

You have got already some good answers.
Here is yet another intuitive way to understand it.
The magnetic field energy carried by the inductor is
$$E=\frac{1}{2}LI^2$$
i.e. it depends directly on the current $I$, not on the voltage $V$.
Due to the energy conservation law this energy,
and hence also the current $I$ "wants" to stay constant.
When you, for example, suddenly open a switch,
and thus interrupt the current $I$, then the inductor
"wants" the current $I$ to continue.
It does so by inducing a high voltage, thus creating
a spark across the opening switch.
This lets the current continue, at least for a short a time.
Then the magnetic energy ($E=\frac{1}{2}LI^2$) is converted
to heat and light in the spark, and finally
the current $I$ drops to zero.
