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So we have an RL DC circuit. If the inductor L1 has a high core permeability, there will be a high inductance and thus a long L/R time constant. So the circuit takes a long time to charge. Would putting a second inductor L2 in parallel with L1 help reduce this time constant at all, if L2 has a much lower core permeability (but is otherwise the same as L1)? If not, is there another way to reduce it other than decreasing L1 or increasing R?

Schematic below. Dotted lines to indicate L2 being added to the circuit. enter image description here

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  • \$\begingroup\$ Welcome to EE.SE! "So we have an RL DC circuit" Please draw a schematic, otherwise it's just guesswork. \$\endgroup\$
    – winny
    Commented Oct 17, 2019 at 10:15
  • \$\begingroup\$ A schematic would greatly help. Have you tried simulating it? LTSpice is great for a quick simulation, and it's free. \$\endgroup\$
    – Puffafish
    Commented Oct 17, 2019 at 10:20
  • \$\begingroup\$ I added a diagram. My computer is very slow right now so I'll have to wait a while to use LTSpice. But good idea, thx \$\endgroup\$
    – Curtwige
    Commented Oct 17, 2019 at 10:30

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Putting another inductor (of any inductance) in parallel with the existing one will reduce the total inductance, so the L/R time constant will be reduced if R stays the same.

If the objective is to get a specified current through a specified inductance and resistance within a short time, increasing the voltage will reduce the time for the current to rise from (say) zero to a given current. It does not change the time constant (which is the time for the current to rise to about 63% of the final current), but it makes the current rise faster.

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