I don't understand the statements for t<0 and t>0. Why does this happen? enter image description here

  • 2
    \$\begingroup\$ Because the switch is closed before t=0, then it is open. \$\endgroup\$ Apr 27, 2016 at 21:10
  • \$\begingroup\$ @BrianDrummond But then R and Ro should carry currents before t<0 \$\endgroup\$
    – Maya
    Apr 27, 2016 at 21:11
  • \$\begingroup\$ What voltage is across them? \$\endgroup\$ Apr 27, 2016 at 21:21
  • \$\begingroup\$ V should be 0 across resistors for current to be 0 ? \$\endgroup\$
    – Maya
    Apr 27, 2016 at 21:33
  • \$\begingroup\$ Good ... now what you need to understand is why V=0. \$\endgroup\$ Apr 27, 2016 at 21:36

1 Answer 1


For t<0, the switch is closed, and all the components are in the circuit. You could think of it as if it was in a steady state, and then I suppose you want to calculate the current in the inductor. Inductors in the steady state behave like short circuits, so all your current will be drawn to it, driving the resistors currents to 0 A. This inductor current (the same as the source, then) is the initial condition for the next step.

When t>0, you should open the switch, and what you have then are 2 circuits: the left one, with source and Ro, and the right one, a RL loop. The calculation for t<0 had the purpose of finding the initial condition for L, since it's current will not disappear instantly when you open the switch. If it has some current, then, it will flow to the resistor and be dissipated, until it disappears.

You see, inductors take a while to "let the current pass" through them, but after that happens, they have no resistance. So, in the dawn of time they behave like open circuits, then they start increasing the current flow through them, until you get to the steady state, when they behave like short circuit.

  • \$\begingroup\$ What I don't understand is why the current through resistors will be zero before t=0? \$\endgroup\$
    – Maya
    Apr 27, 2016 at 21:35
  • 1
    \$\begingroup\$ It's because in a steady state (i.e. after a long time) a inductor behave like a short circuit. I'll edit my answer. \$\endgroup\$ Apr 27, 2016 at 22:08

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