I have been working at this problem for a while.


simulate this circuit – Schematic created using CircuitLab

Given the values of V1 and R1, plus the fact that at the instant SW1 is closed the rate of change of I is approx. 45 A/s, how do I calculate the inductance of L1?

My initial approach was to assume that, because the voltage across the inductor at t=0 is equal to 0v, the back-EMF is equal to 15.5 V, which I can plug into a rearranged emf = -L * (∆I / ∆t), as L = emf / (∆I / ∆t), which yielded a value of 0.34 H. Though this seems to make sense to me, but I'm not sure of this is true due to the fact that I was provided a value for R1.

Let me know if I'm on the right track, I am relatively new to electrical engineering, and might have made some mistakes in my calculations.

I appreciate the help!

edit: as it turns out inductance isn't measured in farads :p

  • 1
    \$\begingroup\$ First hint: Inductance is not measured in farads. \$\endgroup\$
    – John D
    Oct 12, 2021 at 3:59
  • 1
    \$\begingroup\$ The difference between spoon-fed school questions and real life questions is this : in real life you have all sorts of facts, data and measuring equipment available to you, and YOU have to sift out the irrelevant ones and make the relevant measurements. +1 to your school for taking the first step towards that process here. \$\endgroup\$
    – user16324
    Oct 12, 2021 at 12:55
  • \$\begingroup\$ Electron-capture if you are happy with an answer can you formally accept it please. \$\endgroup\$
    – Andy aka
    Aug 14 at 15:12

3 Answers 3


I think you are very much on track to the correct solution. You are using the equations correctly, and are in fact arriving at the correct result. Note, however, that inductance is measured in Henry, not Farad, so the correct answer would be 0.34 H, or 344 mH.

As for the intuition, it is very good that you question why the resistance does not factor into the answer. To see why, consider what happens with the resistor the moment the switch is closed.

Since the inductor prohibits instantanious change in current, the current will start out at zero at the moment the switch is closed. With zero current, the resistor will see zero voltage drop, regardless of what the resistance value is (from ohms law, V = R I). Thus we can conclude that at this instance the value of the resistor is irrelevant.

Of course, once current starts flowing, the resistor will see a voltage drop, and the resistor value will start to matter. If you want to understand the circuit better, try to guess how the graph of current vs. time will look, and perhaps try to figure out how high the current will reach after a long time.


15.5 V / 4.5 ohm = 3.44 A


Here I have 5 volts being applied to a 1 henry inductor (in series with a resistor that is varied from 1 ohm to 8 ohm) at 0.5 seconds: -

enter image description here

Do you see that the initial slope of current for all the graphs is the same?


Well, notice that you've to find (using the initial value theorem):

\begin{equation} \begin{split} \Delta&=\lim_{t\space\to\space0}\frac{\partial}{\partial t}\left(\mathscr{L}_\text{s}^{-1}\left[\frac{\hat{\text{u}}_\text{i}}{\text{s}}\cdot\frac{1}{\text{R}+\text{sL}}\right]_{\left(t\right)}\right)\\ \\ &=\lim_{\text{s}\space\to\space\infty}\text{s}\cdot\text{s}\cdot\frac{\hat{\text{u}}_\text{i}}{\text{s}}\cdot\frac{1}{\text{R}+\text{sL}}\\ \\ &=\frac{\hat{\text{u}}_\text{i}}{\text{L}} \end{split}\tag1 \end{equation}

So, you get:



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