# At $t=0$ the voltage across the Inductor will immediately jump to battery voltage. Why?

"the voltage across the inductor will immediately jump to battery voltage (acting as though it were an open-circuit) and decay down to zero over time (eventually acting as though it were a short-circuit)."

Why the inductor voltage will immediately jump to battery voltage. They say that thats because of the time changing magnetic field across the inductor that will induce equal and opposite voltage in the inductor according to Lenz-Law. But then at t= 0 they claim that the current is zero in the inductor. So when the current is zero in the Inductor then how a magnetic field can be built to induce opposite voltage across the inductor.

So whats happening at t=0 ? zero current and equal and opposite volage ? Since there is zero current then what makes the inductor voltage equal to battery voltage ? If there is current at t=0 then why its never mentioned anywhere in theory or in equations?

And what makes the voltage falls after t=0 after it reachs its maximum value?

• The voltage across the inductor will immediately jump to the negative of the battery voltage. – Myridium Jan 6 '18 at 2:40

The voltage across an inductor is $U_{ind}=L\dfrac{di}{dt}$, it depends not on the current but on its rate of change w.r.t time. Because the current is zero at $t=0$, the voltage across the resistor is also zero (since $U_R=R.i$), Thus, applying Kirchhoff's voltage law, $U_{PN}=U_{ind}$. The voltage across the inductor drops to zero eventually after the current stops growing and reaches a constant value (when $\dfrac{di}{dt}=0$ ).

At t=0 the voltage across the inductor equalizes the battery voltage because by the time instant you start running a current through the inductor, a back EMF of the same value as the battery's voltage is induced, however this induced EMF, according to Lenz' Law, will always oppose the polarity of the battery. Meaning, the inductor at t=0 acts like another battery of the same voltage connected in reverse, hence the voltage across the inductor being equivalent to battery voltage.

Since there is zero current then what makes the inductor voltage equal to battery voltage?

Well that's the point! The reason why there's zero current is because the inductor's voltage is equal to battery voltage even when a series resistor is applied. Which is an equivalent to an ideal open circuit.

The reason why voltage across the inductor falls overtime is because current cannot stay zero. If it did, the changing magnetic field wouldn't be there, therefore no back EMF to restrain the current from running through the inductor. This means that the current will increase overtime -- meaning the inductor will becomes less of an open circuit. Basically the inductor will act like a resistor whose resistance value drops over time. This means that voltage across the inductor drops, until you reach the point where that interpreted resistance drops to zero and your inductor becomes a dead short.

At $t=0$, the electrons rush through the battery to build up on the side of the inductor connected to the negative terminal. Electrons from the other end of the inductor rush through the resistor and to the other side.

The timescale of this process is so much smaller than anything going on with the magnetic field inside the inductor, that we might as well consider it over before the inductor does its work. A quick calculation of the acceleration of electrons affected by a $5V$ field in a $10\,\mathrm{cm}$ wire gives an acceleration of $\approx 8.8 \times 10^{12} \, \mathrm{m} \,\mathrm{s}^{-2}$, from which you can understand why we approximate it as an instantaneous process.