# Transient state analysis of RL circuit

A circuit consists of an ideal battery an ideal inductor and zero resistance. The value of inductance is $$4$$H and the battery has an emf of $$2$$V. There is fuse in the circuit which breaks after when the current in the circuit reaches $$5$$A. Find the time when the fuse blows.

Well, since the circuit has no resistance and the inductor is ideal, the inductor would indefinitely appose the battery with an equal and opposite emf. So, the answer should be $$\infty$$ right? Because the current would remain $$0$$ ALL THE TIME.

But the answer given is $$10$$ seconds.

I then tried applying the KVL to which I got the following equation,

$$L\frac{di}{dt} - 2 = 0$$

Plugging in the values and simplifying I got,

$$\frac{di}{dt} = \frac{1}{2}$$

Considering this equation, the current does reach $$5$$A in $$10$$ seconds. But then, why is my first argument wrong? Doesn't an ideal inductor allow no current and opposes the external emf completely? Or maybe my book is wrong?

Any help would be appreciated.

$$V_{L}(t)=L\frac{di(t)}{dt}$$
This tells you that you can't change the current in an ideal conductor in zero time, because that would mean $$V_{L}$$ would be infinite. But the current can increase gradually over time.
When you switch the inductor onto the battery the current will gradually rise to its maximum value the rate being determined by the inductance and any series resistance that might be in the circuit. However, after a long time in a dc circuit an ideal inductor will look like a short circuit because $$\frac{di(t)}{dt}=0$$ when $$i(t)$$ is constant. This means that unless there is resistance in the circuit an ideal inductor will look like a short circuit across the battery terminals and the current may get very high as it would only be limited by the internal resistance that exists in all real batteries. The fuse in the example limits the maximum short circuit current.