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Assuming this circuit has been running for a long time (switch SW1 is closed), and in t=0 the switch (SW1) opens, and I'm being asked about the current that runs through the inductor L1.

I know that in DC circuit the inductor acts as short and after a very few calculations I came with the current that runs through the inductor to be: 10A.

So I know that for a brief second before the switch opens the current is 10A, how can I make sure or check or explain that the current running through the inductor shortly after the switch opens is still 10A? That there is no "discontinuities" in the current function...

schematic

simulate this circuit – Schematic created using CircuitLab

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2 Answers 2

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... because that's literally the definition of an inductor!

$$v(t) = L \frac{\delta i(t)}{\delta t}$$

This means that for there to be a discontinuity in i(t), its derivative would have to be infinite, and so would the voltage across it. Since the voltage can only assume finite values, i(t) must be a continuous function. You can safely assume that the inductor current just after the switch opens is equal to the value just before it opens. That becomes the starting point for solving the behavior of the circuit from that point on.

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  • \$\begingroup\$ Isn't it possible for the derivative of i(t) to be finite and yet i(t) just after the switch jumps to a different finite value, something like a unit step function? \$\endgroup\$
    – user3921
    Commented May 12, 2014 at 6:21
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    \$\begingroup\$ The derivate is the value of the \$\tan(\varphi)\$ function for the angle of the tangent to the given point. The angle of the tangent for a step would be 90 degrees, what is \$\tan(90^o)\$? \$\endgroup\$
    – WalyKu
    Commented May 12, 2014 at 7:42
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    \$\begingroup\$ @user3921: The derivative of the unit step is the unit impulse, which is a pulse that is infinitesimally narrow and infinitely high. It's a useful theoretical function, but it can't happen in the real world. \$\endgroup\$
    – Dave Tweed
    Commented May 12, 2014 at 10:38
  • \$\begingroup\$ Hey, sorry for the bump. but i got another basic question regarding that situation. How can you explain that the voltage on the inductor is not continuous? it is clear the the voltage shortly before the switch open is 0. but after the switch opens, if i take the derivative of the current that runs through the inductor i get a unit step function (the voltage is 30v shortly after the switch opens), how can i explain that? \$\endgroup\$
    – user3921
    Commented May 12, 2014 at 22:09
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    \$\begingroup\$ @user3921, the slope of the inductor current with respect to time changes from zero to non-zero at the instant the switch is thrown. Since the inductor voltage is proportional to this slope, the inductor voltage changes from zero to non-zero at the instant the switch is thrown. It's just that simple. \$\endgroup\$ Commented May 12, 2014 at 22:38
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how can I make sure or check or explain that the current running through the inductor shortly after the switch opens is still 10A?

It won't be 10A after some finite time elapses after the switch closes.

How to explain this? There are actually two circuits to solve; the circuit before the switch opens and the circuit after the switch opens.

The DC steady state solution for the first circuit yields the initial conditions for the second circuit solution.

So, as you write, the DC steady state voltage across an inductor is zero. Thus, for the first circuit, for \$t=0-\$, the inductor voltage is zero and the inductor current is

$$i_L(0-) = \frac{10V}{1\Omega} = 10A$$

For the second circuit, for \$t > 0\$, we have the following differential equation for the inductor current is (recalling that the fundamental equation for the ideal inductor is \$v_L = L \frac{di_L}{dt})\$:

$$\frac{di_L}{dt} + \frac{R_2}{L}i_L = 0$$

The general solution to this homogeneous, 1st order differential equation is

$$i_L(t) = i_L(0) e^{-\frac{tR_2}{L}}$$

So, we need the initial inductor current to find a particular solution and this is, as stated earlier, the DC steady state solution for the first circuit.

$$i_L(0) = 10A $$

Thus, for any time \$t > 0\$, the inductor current is less than 10A:

$$i_L(t>0) = 10A e^{-\frac{tR_2}{L}} < 10A $$

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  • \$\begingroup\$ I like all of your answers, Alfred because they are normative. \$\endgroup\$
    – Roh
    Commented May 12, 2014 at 11:13

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