I was given the following problem in a textbook. My teacher reviewed the problem for the class and I was presented with, what I suspect is, a wrong answer. I am asking here to see if my own reasoning is correct.

The problem:


simulate this circuit – Schematic created using CircuitLab

In the circuit above... The Inductor has an internal wire resistance of 0.5 Ohms. The resistor is 300 Ohms. Before throwing the switch, there is 2 Amps through the inductor. What is the "maximum induced voltage" of the inductor. (The wording and values have been changed to protect the not so innocent.)

My solution:

My assumptions:

  • Ideal components besides the resistance of the inductor
  • Conventional current flow and clockwise

My math: \begin{eqnarray*} V_R + V_L = 0\\ (I\cdot R_R) + (L\cdot dI/dt + I\cdot R_L) = 0\\ 2\cdot 300 + (stuff + 2\cdot 0.5) = 0\\ stuff=-601\\ V_L = (stuff + 2\cdot 0.5) = -600 volts \end{eqnarray*} The internal wire doesn't affect the answer to this idealized construction, right?

  • 1
    \$\begingroup\$ I'm curious: What was the supposed wrong solution of your teacher and how did he explain it? \$\endgroup\$
    – Elec1
    Aug 19, 2021 at 7:44
  • \$\begingroup\$ It wasn't much of an explanation as just V=IR=2*(300+0.5)=601v. So I guess I got hung up on wording and not understanding what they were asking for. EMF around the loop verses max voltage across the inductor \$\endgroup\$
    – 9Harris
    Aug 22, 2021 at 1:11

1 Answer 1


The maximum induced voltage across the inductor terminals will be 600V.

One can easily arrive at the result by noting that the current through the inductor just prior to throwing the switch was 2A. Once the switch is thrown, that 2A needs to flow through the 300\$\Omega\$ resistor. V=IR = 2A x 300\$\Omega\$ = 600V.

However, one should include the 0.5\$\Omega\$ in the calculation of the total induced emf (i.e. the emf around the loop). The inductor's internal resistance adds another 1V. Although that 1V is dropped by the resistance of the coil, it is still "induced" voltage, because the total current around the loop times the total resistance around the loop must equal the total emf around the loop. So the peak emf around the loop is 601V.

  • \$\begingroup\$ That makes sense! I guess I was hung up on thinking about where I could measure the induced voltage. Wording it as looking for the peak "emf" helps me consider the 1v more. It's the "extra 'force'" to over come the extra 0.5 ohms. \$\endgroup\$
    – 9Harris
    Aug 22, 2021 at 1:06
  • \$\begingroup\$ So from the wording of "maximum induced voltage by the inductor", would you call it 601v because it's the voltage the inductor creates to over come both the resistor and its internal resistance? \$\endgroup\$
    – 9Harris
    Aug 22, 2021 at 1:13
  • \$\begingroup\$ @9Harris I wouldn't quite say the voltage "overcomes" the resistance. I might say that the 601v is the voltage the inductor creates in order to maintain a 2 amp current through the 300.5 ohm total resistance. \$\endgroup\$ Aug 22, 2021 at 2:13

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