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

  • \$\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


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

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

  • \$\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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.