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real circuit

circuit after t0+

When 225 mA comes to the top node, why doesn't all of it go through the short circuit? I think 160 mA on the inductor is what is left from t0- and shouldn't be considered in the equations anymore. I know the solution uses KCL to provide the answer but if only 65 mA goes through the short circuit, where does 160 mA leftover goes? The only possibility I see is that it goes through the inductor but if so current on the inductor at t0+ would be 320 mA or it will have a current of 160 mA as t goes to infinty which is a contradiction to inductor fundamentals.

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  • \$\begingroup\$ I(120R)=65mA, ia=io \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 2 '17 at 21:44
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    \$\begingroup\$ There's no short circuit anywhere.. \$\endgroup\$ – Harry Svensson Dec 2 '17 at 21:50
  • \$\begingroup\$ The Op has mistaken the current paths \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 2 '17 at 22:09
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    \$\begingroup\$ what is top node? there are several nodes in the circuit. specify which node you are referring to. \$\endgroup\$ – jsotola Dec 2 '17 at 22:10
  • \$\begingroup\$ inductors resist changes in current ... you are calculating currents at the moment when the switch closes ... for an infinitesimally short time after the switch closes, the inductor will maintain the same current flow \$\endgroup\$ – jsotola Dec 2 '17 at 22:21
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If there was a steady state achieved before closing the switch, The inductor current goes on just like already commented. Inductor current will decay (=decrease) with time constant L/R = 100mH/20Ohm = 5ms after the switch is closed.

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