# Find voltage across inductor in RL circuit

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.

• Here you have a tip electronics.stackexchange.com/questions/392657/…
– G36
Jan 18, 2020 at 12:23
• Don't you need to find $I_{R2}$ at $t_{0-}$ (before the switch is opened) first? Jan 18, 2020 at 12:26