Find voltage across inductor in RL circuit

Question

I have this circuit in wich I try to find the voltage across the inductor (VL) at t = 0+ and the switch OPENS at t = 0

^{simulate this circuit – Schematic created using CircuitLab}

Here are my calculations below, but I can't seem to get a correct answer. Can anyone help me? or tell me how to do it?

I know that if t tends to infinity, the inductor acts as an short circuit, which allows me to calculate the equivalent resistor: $$R_{eq} = 3.75 \mbox{ kOhm}$$ ((4+2)//(6+4)) and the general voltage when the switch is open: $$V=3.75*10=37.5V$$

So I have $$I_{R1} = 6.25 mA$$ and $$I_{R2} = 3.75 mA$$

And I know that on resistor R2, the current is $$\frac{V_L-V}{6} = 3.75$$ wich gives me $$V_L = 3.75*6+V = 3.75*6+37.5 = 60V$$

I do not have the answer but I know that this answer is not correct (we submit our responses to an online quiz which is corrected directly). Is there an error in what I did?

Thanks a lot.

Answer 1

If I told you that there is 5 mA flowing through the inductor just before the switch opens can you see that? It’s based on the assumption that the switch has been closed for a long time. Do you understand that bit?

Can you also see that as the switch opens, in that instant there will still be 5 mA flowing in the inductor?

Based on that, you can calculate the voltages across R2 and R3 and the current source. The voltage across the current source is defined, in that instant, by 5 mA also flowing through R1 and R4.

Can you take it from here?