In the below example, why is the voltage source shorted for t>0, and why does it say "the 6-ohm resistor is shorted" in the paragraph about t<0? Isnt it open circuit there?

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    \$\begingroup\$ When we say the switch was open "for a long time", we mean long enough for the circuit to reach its DC steady state operating point. How do we analyze an inductor in DC steady state? \$\endgroup\$ – The Photon Sep 4 '18 at 3:25

At a steady state (DC conditions), the inductor will have no resistance, and will act like a short. Since current wants to take the path of least resistance, all the current will flow through the inductor (instead of the 6 ohm resistor).

For t<0, since the inductor has 0-ohms of resistance, it's short circuiting the current path through the 6 ohm resistor.

For t>0, once the switch closes, all the current from the voltage source will take the path of least resistance through the closed switch leg (as opposed to the other leg that it is in parallel with).

Sorry if I went very basic with this explanation. Hope it helps :)

  • \$\begingroup\$ for t<0, im confused by the terminology. If the current doesnt go through the wire containing the 6-ohm resistor, isnt that acting like an open circuit? Also, for t>0, how is it physically possible for the left side of the circuit to have its own current running separately while at the same time have current from the inductor take effect? \$\endgroup\$ – David Sep 4 '18 at 3:56
  • \$\begingroup\$ @David Think about it. If there is an ideal inductor (zero Ohms, some non-zero inductance) in parallel with a resistor, which direction will the current go? Through \$6\:\Omega\$? Or through the inductor's \$0\:\Omega\$? It's all going to go through the inductor, given enough time to overcome the inductance's reluctance to allowing changes in current. The inductor, given infinite time, looks like a dead short across the \$6\:\Omega\$ resistor. All the current will go through the "short" and not through the resistor. \$\endgroup\$ – jonk Sep 4 '18 at 4:38
  • \$\begingroup\$ @David That said, once the switch is flipped, the current through the inductor cannot be instantly changed, either. So the inductor's current starts out at the same value it was at when the switch flipped. Now, that current must either go through the \$6\:\Omega\$ resistor or else the \$3\:\Omega\$ resistor or else both of them. In this case, it is pretty obvious that about twice the proportion will go through the \$3\:\Omega\$ than through the \$6\:\Omega\$. So it should be about 2/3 amp through the \$6\:\Omega\$ and about 4/3 amp through the \$3\:\Omega\$.... at t=0. After that, it changes. \$\endgroup\$ – jonk Sep 4 '18 at 4:42

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