when an RL circuit is opened for a long time then closed at $t=0$, why is there a sudden jump in its change in current ($\frac{dI}{dt}$) the moment the switch is closed? And why is this value so large?
Is this something to do with the resistance of the inductance trying to not allow any current flowing through when a battery is initially connected?
So imagine it was open like so for a long time, then at $t=0$, the swtich closes connecting the battery to the rest of the circuit.
(Almost) instantaneously the switch is closed the potential difference across resistor $R_2$ becomes $\dfrac{R_2}{R_1+R_2}\cdot E$ - potential divider.
Before the switch was closed the current through the inductor was zero.
When the switch is closed the current through an inductor is zero as the current cannot change instantaneously.
The induced emf produced by the inductor is $\dfrac{R_2}{R_1+R_2}\cdot E = L\,\dfrac{dI}{dt}$ and the induced emf is produced because the current is changing.
As to whether that rate of change of current is large depends on what you mean by large.
Here is the circuit simulated in the Spinning Numbers Sandbox.
To simulate the switch the input voltage is a step of $1\,\rm V$.
The red voltage probe measures the potential difference across nodes $a$ and $b$ and the blue current probe measures the current flowing through the inductor.
The response of the circuit is selected by clicking on the $\fbox{TRAN}$ button and is over a period of $10\,\rm ms$.
The time constant of the circuit is $2\,\rm ms$ with the initial voltage of node $a$ being $0.5\,\rm V$ and dropping towards zero volts over time.
The initial current through the inductor is zero and increasing towards $1\,\rm mA$ over time.
This is consistent with the idea that the inductor has zero resistance and no emf is generated when the current is not changing.
Values of voltage $452 \,\rm mV$ and current $95 \,\rm \mu A$ after $200\,\rm \mu s$ are shown on the graph.
So the initial rate of change of current is $\dfrac {95 \,\rm \mu A}{200\,\rm \mu s}\approx 0.5\,\rm A/s$.
Remembering that the induced emf is $L\dfrac{dI}{dt}$ the rate of change of current is approximately $\dfrac{0.5\,\rm V}{1\,\rm H} = 0.5\,\rm A/s$
Finally my first two words, (Almost) instantaneously, refer to the fact that in the real world nothing happens instantaneously and the potential divider voltage reaches a value of $0.5\,\rm V$ after a short span of time dictated by stray/parasitic inductance (see the loops) and capacitance in a real circuit.
Heuristically, at very short times an inductor acts as an open circuit and at very long times an inductor acts as a short circuit.
So as soon as you close the switch the inductor acts as an open circuit so all of the current goes through the resistors. That makes the voltage across the 2nd resistor immediately jump. Then, because the inductor is in parallel with the 2nd resistor, that same voltage is across the inductor. A voltage across an inductor is proportional to the rate of the change in current through the inductor.
So the reason that the $dI/dt$ suddenly jumps is because the voltage suddenly jumps and $dI/dt$ is proportional to $V$.
