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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

schematic

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.

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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?

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  • \$\begingroup\$ I see the firts part: Req = 3k so V = 30V and the current trough the inductor is 5mA. I would say that as the swith opens the current trough the inductor decreases slowly so when the switch is open there is still 5mA in the inductor. And the voltage across R2 is 30V right ? So I only have 7.5 volts left after R2. \$\endgroup\$
    – Pierre
    Jan 18, 2020 at 12:59
  • \$\begingroup\$ Yes, with 30V across R2 the left hand side of the inductor is at 0V. Now there is still 5mA flowing in R3....... \$\endgroup\$
    – James
    Jan 18, 2020 at 13:20
  • \$\begingroup\$ @Pierre yes that's quite correct and, moving on, we know that the voltage across the current source will remain at 30 volts at the instant the switch opens because there will still be 5 mA flowing into R1 and R4. Given that there has to be 30 volts across R2 then the left side of the inductor MUST drop to 0 volts (as mentioned by James). Have you got your answer yet? \$\endgroup\$
    – Andy aka
    Jan 19, 2020 at 12:54
  • \$\begingroup\$ Yes I get it, thanks for the help ! \$\endgroup\$
    – Pierre
    Jan 21, 2020 at 16:54

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