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So I'm working on an RL circuit and the solution is stated to be that R1 and R3 are in series, but I didn't think that was the case because they don't have the same current through them. Given solution says to add R1 + R3 in series and then use that equivalent resistance and add to R2 as if it were in parallel.

TL;DR: How do I find the equivalent resistance of the circuit below so that I can find the time constant for this RL circuit?

Circuit

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    \$\begingroup\$ In order to add R1+R3 in series you'll have to transform I1 into a voltage source. So I1 becomes say V1 with 600V and R1 is now in series with R3. See source transformation \$\endgroup\$ – dopamane Sep 28 '18 at 0:33
  • \$\begingroup\$ Thevenin's Theorem says this voltage can be zero. For DC resonance I1 can be omitted because it does not change the RL resonance value as it is not a 'frequency'. Add R1+R3 in series then add R2 in parallel. Now you have your RL values. \$\endgroup\$ – Sparky256 Sep 28 '18 at 4:40
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If you are trying to find the R-L time constant (L/R numerically) then the constant current generator has infinite impedance and can be ignored. This leaves R1 and R3 in series (R1+R3) in parallel with R2 hence the effective resistance of the circuit is: -

R2 || (R1 + R3)

Or do it using Thevenin's theorum; a current source in parallel with a resistor (R1) is equivalent to a voltage source in series with that resistor and this of course makes R1 in series with R3.

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