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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

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  • \$\begingroup\$ "Any input would be much appreciated" -->Input ---> draw a better circuit that is legible. \$\endgroup\$
    – Andy aka
    Commented 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\$ Commented Feb 12, 2016 at 13:09
  • \$\begingroup\$ Updated the circuit, sorry. \$\endgroup\$ Commented 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
    Commented 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\$ Commented Feb 12, 2016 at 17:50

1 Answer 1

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schematic

simulate this circuit – Schematic created using CircuitLab

Each resistor will have 1A current flow.

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  • \$\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\$ Commented 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\$ Commented 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\$ Commented 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\$ Commented 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\$ Commented Feb 12, 2016 at 14:15

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