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TL,DR: Why does i_L in this point down instead of up? I get two different answers for the current i_2 when I imagine them in these different ways. Or am I (likely) making a mistake or missing some finer point here?

Figure 8.21


This is an image from an example question in my Introduction to Circuits textbook. For the sake of helping other users find this, I'll quote the text here.

8.5 At t = 0.15 s in the circuit of Fig. 8.21, find the value of (a) i L ; (b) i 1 ; (c) i 2 .


Ans: 0.756 A; 0; 1.244 A

As I picture it,

  • The current coming from the 2 A current source at t > 0 flows upwards, to the right, and then immediately downwards and back to the source again.
  • i_L flows downwards, bypasses the 8 ohm resistor, and then upwards "against the stream" of the 2 A source, and then returns to the inductor through the 2 ohm resistor.

But I'm unclear as to why i_L has to flow downwards in this case. It seems to me it would be just as correct to have i_L start upwards, flow through the 2 ohm resistor first, and then downwards through the short circuit instead. It's the same path, after all.

Is there a reason my textbook wrote that i_2 should be the difference between 2 A and i_L?

Thank you very much for your time.

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  • \$\begingroup\$ Is the question to find the steady-state currents? Also current arrows are arbitrary, if it actually flows the opposite way to the direction assumed it will just come out negative. Sorry, realized it specified the time in the question. \$\endgroup\$ – jramsay42 Feb 3 '18 at 7:01
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Before closing the switch:

  • Initially inductor was an open circuit. At the steady state it can be assumed as a short, and a steady state current \$I \$ will flow "downwards", through 2 ohms and the inductor. The value of it will be \$ 2* \frac {8}{(2+8)} \$ = 1.6 A.

After closing the switch:

  • The inductor will act like a current source and it sources a current \$ I_L\$ = 1.6 A at t = 0 . "Downwards" itself. Because an inductor opposes any change in current through it . An emf is induced, and as per lenz' law, the effect of the induced emf in L will be to keep the current flowing in the same direction.
  • Equivalent circuit after shorting:

    enter image description here

Apply KCL at any node where currents meet. So at any moment t > 0 ,$$i_2 + i_L = 2$$ $$ i_ 2= 2 - i_L$$

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