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I want to ask that how to get the time constant for simple RCL circuit in order to get the natural response later, Here is the sample circuit. Assume that S is close at t=0 and the circuit is open for a long time

schematic

simulate this circuit – Schematic created using CircuitLab

when S close, in order to calculate the time constant \$\tau\$ in this case, I omit the source \$V_S\$ but then I don't know how the circuit which will be used to calculate \$\tau\$ look like. I thought \$V_S\$ will act like sort circuit so it'll look like this

schematic

simulate this circuit

but my lecturer said it actually looked like this, which doesn't have \$R_1\$

schematic

simulate this circuit

So what is the rule when omitting the source in case of: voltage source and voltage source. What if the switch is in series but not parallel?

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  • \$\begingroup\$ No particular rule - just look at the circuit harder and notice that R1 is totally bypassed when switch closes. Use your eyes and redraw correctly. \$\endgroup\$ – Andy aka Oct 14 '15 at 12:36
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I thought \$V_S\$ will act like short circuit

This is why you're confused.

In this circuit, we don't have any control over \$V_S\$ - it is a constant, unchanging, DC voltage. However, we do have control over the switch in the circuit. When, we close the switch, the circuit splits into two separate pieces.

The left side of the circuit becomes:

schematic

simulate this circuit – Schematic created using CircuitLab

Since this circuit has nothing to do with the inductor, we can ignore it.

The right side of the circuit is what your teacher drew, so we can simply analyze that on its own. Note that \$V_S\$ hasn't changed - instead, we've totally taken it out of the circuit.

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  • \$\begingroup\$ hi, how about when the switch is in series and DC and \$R_1\$ are involved? Can I omit the source and treat it like short circuit or what? \$\endgroup\$ – aukxn Oct 14 '15 at 13:13
  • \$\begingroup\$ Notice that the switch does not magically turn off the DC source - in this case, it just separates it from the inductor's circuit. If the switch was in series, the DC source would be connected to the rest of the circuit, of course. I think you need to take a step back and focus on what the circuit really looks like with the switch opened and closed instead of learning specific "in series/parallel" cases. \$\endgroup\$ – Greg d'Eon Oct 14 '15 at 13:16

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