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I don't know if it's just me but I can't find a single resource that derives a general equation for the current through an inductor. My text book just says "assume the equation if of the form ..." but I want to understand the assumption comes from and be able to construct and solve the differential equation if I ever forget the form.

If possible could someone derive the equation similar to the way this person derived the equation for a capacitor?

Deriving the formula from 'scratch' for charging a capacitor

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  • \$\begingroup\$ How familiar are you with magnetism and Maxwell's equations? Start from there, and assume a coil of wire around an infinitely permeable core with a small air gap, and see what relationships you can find between \$i\$, \$λ\$, and \$v\$. \$\endgroup\$ – Hearth Nov 8 '18 at 3:44
  • \$\begingroup\$ @Felthry I think the OP was asking a question that skips past the development anyone can find here. I suspect it was more about applying a simple differential equation, instead. Nothing fancier, my guess, given the link the OP provided. \$\endgroup\$ – jonk Nov 8 '18 at 5:18
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schematic

simulate this circuit – Schematic created using CircuitLab

Like in the case of the capacitor, there are three steps:

  1. Write a KVL equation which will lead to a differential equation.
  2. Solve the differential equation.
  3. Apply the initial condition of the circuit to get the particular solution.

The KVL equation can be written simply as

$$ V_i = v_L + v_R $$

The equation for an inductor is slightly different from a capacitor however. Here, the flux is given by \$\Phi = L\cdot i\$, and the EMK generated by this flux is \$v_L = \frac{d\Phi}{dt} = L\frac{di_L}{dt}\$. Alternatively, you can also write that

$$ i_L = \frac{1}{L}\int_0^t v_L(u)du + i_L(0) $$

The resistor and the inductor share the same current, so

$$\begin{align} i_R &= i_L \\ &\Downarrow \\ v_R &= R\cdot i_R = R\cdot i_L \\ &\Downarrow \\ v_R &= \frac{R}{L}\int_0^t v_L(u)du + R\cdot i_L(0) \end{align}$$

We can put this into the KVL equation:

$$\begin{align} V_i &= v_R + v_L \\ &\Downarrow \\ V_i &= \frac{R}{L}\int_0^t v_L(u)du + R\cdot i_L(0) + v_L \\ &\Downarrow \\ V_i - R\cdot i_L(0) &= \frac{R}{L}\int_0^t v_L(u)du + v_L \end{align}$$

We can find the solution by first integrating both sides of the equation.

$$ \frac{d(V_i - R\cdot i_L(0))}{dt} = 0 = \frac{R}{L}v_L + \frac{dv_L}{dt} $$

We can work this out as follows:

$$\begin{align} -\frac{R}{L}v_L &= \frac{dv_L}{dt}\\ &\Downarrow\\ -\frac{R}{L}dt &= \frac{dv_L}{v_L}\\ &\Downarrow\\ -\frac{R}{L}\int_0^tdt &= \int_{v_L(0)}^{v_L(t)} \frac{1}{v_L} dv_L\\ &\Downarrow\\ -\frac{R}{L}t &= \left[ \ln(v_L) \right]_{v_L(0)}^{v_L(t)}\\ &\Downarrow\\ -\frac{R}{L}t &= \ln\left(\frac{v_L(t)}{v_L(0)}\right)\\ &\Downarrow\\ v_L(0)e^{-\frac{R}{L}t} &= v_L(t) \end{align}$$

\$v_L(0)\$ is the voltage across the inductor at \$t=0\$, which is equal to \$V_i - R\cdot i_L(0)\$.

So if the inductor starts out discharged, ie. \$i_L(0) = 0\$, then \$v_L(0) = V_i - R\cdot i_L(0) = V_i\$, and the fomula will turn into

$$v_L(t) = V_ie^{-\frac{R}{L}t}$$

If the inductor already had a current before, and \$V_i\$ is suddenly turned to 0, then \$v_L(0) = 0 - R\cdot i_L(0) = -R\cdot i_L(0)\$ and the solution will turn into

$$v_L(t) = -R\cdot i_L(0)\cdot e^{-\frac{R}{L}t}$$

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  • \$\begingroup\$ Personally I prefer using the Laplace transformation, or in this case working to \$i_L\$ to avoid the integral in the differential equation. I opted to follow the same procedure as your example post since that seemed to be your question. \$\endgroup\$ – Sven B Nov 8 '18 at 12:35
  • \$\begingroup\$ A lot of extra math could have been removed. Just set up the nodal, take its derivative, rearrange and integrate, and solve the constant of integration at t=0. Done in 5 or 6 steps at most. But right, of course. \$\endgroup\$ – jonk Nov 8 '18 at 13:45
  • \$\begingroup\$ @jonk I tried to follow the structure of the accepted answer of the linked example in the question, which also (I assume deliberately) included more steps than necessary. \$\endgroup\$ – Sven B Nov 8 '18 at 14:05
  • \$\begingroup\$ I gave a +1. But I'm not sure the OP is prepared for a more abstract approach. It is what it is, though. Hopefully, just fine. \$\endgroup\$ – jonk Nov 8 '18 at 14:07
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A useful 'trick' to give the output function, \$\small f(t)\$, of a 1st order system subject to a step change of input function, is:

$$\small f(t)= final \:value + (initial \:value - final \:value)e^{-t/\tau}$$ or $$\small f(t)= f(\infty) + (f(0) - f(\infty))e^{-t/\tau}$$.

The initial and final values, and the time constant can usually be found by inspection.

In this particular case the initial value of current is \$\small i(0)=0\$; the final value is \$\small i(\infty)=\frac{V_i}{R}\$; and the time constant is \$\small \tau = \frac{L}{R}\$. Hence $$\small i(t)= \frac{V_i}{R} + (0 -\frac{V_i}{R})e^{-Rt/L}$$

or $$\small i(t)= \frac{V_i}{R}(1 -e^{-Rt/L})$$

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