So, I have to find the initial conditions for the following circuit with a switch. For \$t<0\$, the switch is closed, so we draw the inductor as a short circuit and the conductor as an open circuit. I have to find the voltage drop across the conductor, which would be equal to the voltage drop across the \$R_2\$ rezistor, as well as the current through the inductor.

Here comes the problem, since the current source \$J_2\$ is short circuited, how would I go about solving this? From what I know, the source wouldn't be eliminated and the current flowing through the short circuit would be equal to the source, \$i_L(0-) = 4 A\$. However, I'm not sure how I could go about finding the current flowing through the \$R_2\$ rezistor, as the current source parallel to the short circuit is confusing me.

Any input would be much appreciated. Thanks in advance!

enter image description here

  • \$\begingroup\$ "Any input would be much appreciated" -->Input ---> draw a better circuit that is legible. \$\endgroup\$
    – Andy aka
    Feb 12, 2016 at 13:06
  • \$\begingroup\$ Replace R3 and I1 with thevenin equivalent, 6V + 3ohms and now you just have R1, R2 and R3 in series with 6V source. \$\endgroup\$ Feb 12, 2016 at 13:09
  • \$\begingroup\$ Updated the circuit, sorry. \$\endgroup\$ Feb 12, 2016 at 13:40
  • \$\begingroup\$ Which switch? There is no switch in your circuit. What exactly happens at t=0? Drawing a circuit like this assumes that it has been in this state since the beginning of time, therefore it has reached it's steady state a long time ago. \$\endgroup\$
    – Bart
    Feb 12, 2016 at 15:55
  • \$\begingroup\$ I forgot to add the switch when I built the digital schematic. There is a switch next to the R2 rezistor which opens at t=0. \$\endgroup\$ Feb 12, 2016 at 17:50

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Each resistor will have 1A current flow.

  • \$\begingroup\$ I have no problem solving this circuit but what confuses me is how you got this. Is the second source, \$I_2\$, completely ommited? Thanks for your reply? \$\endgroup\$ Feb 12, 2016 at 13:27
  • \$\begingroup\$ Yes, the inductor is a short circuit at DC. I2 will flow through the inductor, but it will produce no volt drop and have no effect on the rest of the circuit. \$\endgroup\$ Feb 12, 2016 at 13:41
  • \$\begingroup\$ So, if we want to measure the current flowing through the short circuit, it would be equal to \$I_2 = 4 A\$, right? So, how come, through your analysis, the current flow through the short circuit results \$1 A\$? Shouldn't we take into consideration the current flowing from the second source? \$\endgroup\$ Feb 12, 2016 at 13:48
  • \$\begingroup\$ I said all the resistors will have 1A flowing in them, and they will. The only place that I2 make a difference is in L1. L1 will have 4A from I2 plus the 1A flowing in R1,R2 = 5A. \$\endgroup\$ Feb 12, 2016 at 13:55
  • \$\begingroup\$ Okay, thanks for the clarification, this was the part that was confusing me. Now it makes sense! \$\endgroup\$ Feb 12, 2016 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.