A capacitor of 1.58mF was charged by a battery of 12V and lies on the ground of an abandoned radio station. During a storm, an old telephone cable fell from a rack and one of its tip got in touch with one terminal of the capacitor. The telephone cable has a resistance of 14mΩ and an inductance of 15µH.

A wet racoon entered the building and accidentally stepped on the not connected terminal of the capacitor and the telephone table. Knowing that he suffered 18µs before being able to run away and that the current that made him squirm was of 100mA, what is the value of the resistance of the racoon?


simulate this circuit – Schematic created using CircuitLab

I imagined the circuit as above, where R is the resistance of the racoon. I supposed 18µs was to be considered as instantly after 0, so I had the initial value of current to find the coefficients with: $$ \frac{di(t)}{dt} = - \frac{1}{L} \cdot (RI_o + V_o)$$ where $$ I(0^+) = I_o = 100\,\,mA $$ and $$ V(0^+) = V_o = 12\,\,V$$

From this, I tried to solve the EDO bellow:

$$ i''(t) + \frac{R_{eq}}{L} + \frac{i(t)}{CL} = 0$$ where $$ R_{eq} = R_{racoon} + 14 \cdot 10^{-3}$$

But I'm not able to go further, since it seems not possible to know its root without R, nor its coefficients (bellow is the equation I found manipulating the variables):

$$-\alpha\,\pm \, \sqrt{\alpha^2 - w_o^2}\,\, =\,\, \frac{R_{eq} \, \pm \,\sqrt{C^{-1}(R_{eq}^2C-4L)} }{2L} $$

So I kept thinking, is this question even analytically possible?

  • \$\begingroup\$ Why is there a short across your voltage source? \$\endgroup\$
    – Andy aka
    Commented Apr 30, 2018 at 8:38
  • \$\begingroup\$ @Andy aka The Voltage source has a step unit in it that will cancel out its value when t = 0, leaving only the RLC circuit above, in which the capacitor was fully charged at t = 0 and begins to discharge when the racoon connects the circuit with his wet paws. \$\endgroup\$ Commented Apr 30, 2018 at 10:00
  • \$\begingroup\$ I'll try again........ Why is there a short across your voltage source? \$\endgroup\$
    – Andy aka
    Commented Apr 30, 2018 at 10:20
  • \$\begingroup\$ This question can not be answered without knowing the resistance (ESR) of the capacitor. Imagine a capacitor with huge resistance: It will never drive 100mA into the load. I assume the story teller is thinking of an ideal capacitor? (Normally, an 1.58mF capacitor has a big resistance.) \$\endgroup\$ Commented Apr 30, 2018 at 10:31
  • \$\begingroup\$ Does "lies on the ground of an abandoned radio station" imply that one side of the capacitor is at least somewhat connected to this ground? This would not be the case for ordinary electrolytic capacitors since they have a plastic jacket around them. \$\endgroup\$ Commented Apr 30, 2018 at 11:28

2 Answers 2


Some approximations might be helpful.

100mA after 18us means, \$C\frac{dV}{dt} = 100mA\$ $$\implies dV \approx 1mV$$ The voltage of the capacitor has therefore lowered only very little in that period ( 12V to 11.999V). Hence the circuit maybe considered like a constant DC voltage source driving a RL circuit. The current equation is therefore -

$$i.e., \frac{12}{R}(1-e^{-t\frac{(R+14\times10^{-3})}{L}})=100mA$$ t, L are known.

NOT so easily solveable though.

  • 1
    \$\begingroup\$ I think that this equation is not analytically solveable (At least not "easily" for me ;-) \$\endgroup\$ Commented May 2, 2018 at 4:57
  • \$\begingroup\$ Sorry, but I struggle to see how this is easy too. I don't know how to isolate R to compute the value. Could you show us how you would do it? \$\endgroup\$ Commented May 2, 2018 at 15:21
  • \$\begingroup\$ @stefan lol :-D I just realised that. Will end up in an equation of form : mR + ne^R= k where m,n,k are constants we have to solve for R. Interestingly it looks easy, but actually soo complex to solve analytically. Would be a math stack exchange puzzle. \$\endgroup\$ Commented May 2, 2018 at 17:14
  • \$\begingroup\$ ;-( edited..... \$\endgroup\$ Commented May 2, 2018 at 17:15
  • \$\begingroup\$ You can solve this graphically though. Plotting the corresponding graphs using wolfam online tool. And find the meeting point. It would be the solution. \$\endgroup\$ Commented May 2, 2018 at 17:18

Here’s a plan:

The capacitor can be dismissed. The voltage drop at 18us is very tiny because of the huge value of the cap - so you can replace it with constant U0=12V in the model. So there's only a RL circuit left.

The solution to the differential equation of the RL circuit with starting voltage U0 is: I(t)=U0/R*(1-exp(-tR/L)).

I don't know how to solve this (analytically) for R, but asking wolfram alpha, it gives me a result of about 120 Ohms. The racoons resistance is 119.986 Ohms.


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