I have a problem with this circuit. I am supposed to calculate i(t) at the moment t=(0+). The parameters are:
Switch is closed at t=0

I know that current can't jump in the branch with the inductor, so


but I'm not sure what will happen with the rest of the circuit. Is there a simple way to get i(t)?
Sorry about quality of the picture.


  • \$\begingroup\$ Is there a simple way to get i(t)? Yes, actually :) Kirchoff's Laws. Since you mentioned, I'm not sure what will happen with the rest of the circuit, I assume that you might not know Kirchhoff's Laws. \$\endgroup\$
    – user103380
    Sep 8, 2019 at 20:36
  • \$\begingroup\$ KingDuken would you mind telling me, if it's so simple? I do not have a problem with applying Kirchhoff's laws, it's more that I do not know what will happen with middle branch of the circuit. Will the currrent through middle resistor be 0 or will it jump? I realize that i(t) is a sum of currents in other branches. \$\endgroup\$
    – Harvastum
    Sep 8, 2019 at 20:53
  • \$\begingroup\$ The schematic shows a current source, but fails to define the polarity of the voltage e(t) across it... \$\endgroup\$
    – Huisman
    Sep 8, 2019 at 21:02
  • \$\begingroup\$ @Harvastum I would recommend using mesh current loops. Your inductor will act like a short circuit, when you use this method. \$\endgroup\$
    – user103380
    Sep 8, 2019 at 21:04
  • \$\begingroup\$ There's an arrow pointed upwards. (the + would be up) \$\endgroup\$
    – Harvastum
    Sep 8, 2019 at 21:04

1 Answer 1


You are correct that the current in the inductor can't change instantly. But it certainly can in any of the resistors. So, immediately after the switch closes, you have this situation:


simulate this circuit – Schematic created using CircuitLab

where I1 is the same as the coil current before the switch closed. Now, what's the current through V1?

  • \$\begingroup\$ Thank you so much! This analogy is exactly what I needed. I've got 30mA by the way. \$\endgroup\$
    – Harvastum
    Sep 8, 2019 at 21:56
  • \$\begingroup\$ Note that this is essentially the same method you use for "right after switch closure" when capacitors are involved -- except that they become voltage sources, not current sources. \$\endgroup\$
    – TimWescott
    Sep 9, 2019 at 2:37

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