I derive the formula for energy stored in an inductor as follows:

$$\text{power} = P = v(t)i(t)$$

$$\text{energy} = E(t) = \int \limits_{t_0}^{t} P(\tau)~ d\tau$$

$$\text{energy} = E(t) = \int \limits_{t_0}^{t} v(\tau)~i(\tau) d\tau$$

substituting voltage law for inductor:

$$v(t) = L \frac{di}{dt}$$

I get:

$$E(t) = \int \limits_{t_0}^{t} L \frac{di}{d\tau}~i(\tau) d\tau$$

$$E(t) = L \int \limits_{i(t_0)}^{i(t)} I ~dI$$

$$E(t) = \frac{L}{2} \bigg[I^2\bigg]^{i(t)}_{i(t_0)}$$

$$\boxed{E(t) = \frac{L}{2} \bigg[i^2(t) - i^2(t_0)\bigg]}$$

However the book that i'm using defines inductor energy as just:

$$E = \frac{1}{2}~L~i^2~~[\text{joule}]$$

then they go onto to calculate the energy of a 2 Henry inductor with current flow of :

$$i(t) = 2e^{-t}$$


$$E = \frac{1}{2}(2)(2e^{-t}) = 4 e^{-2t}~~[\text{joule}]$$

Strictly speaking, is this right??

If I consider the current flow at time zero t=0, it has an initial current of 2 amps... and I don't see that factored into their energy equation. Or did I do something wrong in my derivation of the inductor energy equation and it shouldn't have the i(t_0) term?

I suppose i could play a trick like this by defining current like this:

$$i(t) = \begin{cases}2e^{-t} & t\ge 0\\0 & t \lt 0\end{cases}$$

but, it doesn't really help since the initial current is still 2 amps. and you really can't jump an inductor current instantaneously...

  • \$\begingroup\$ Surely at time = 0 the current is 0, then at time 0 + x (a small time) the current is starting to flow... \$\endgroup\$
    – Solar Mike
    Jan 10, 2020 at 15:59
  • \$\begingroup\$ $$i(t=0)=2e^{-0} = 2~A$$. No? they didn't really speciy that current was non-zero before t=0. \$\endgroup\$
    – pico
    Jan 10, 2020 at 16:00
  • \$\begingroup\$ Perhaps the book uses the energy at t = 0 as a reference, they care only about the energy difference. \$\endgroup\$ Jan 10, 2020 at 16:01
  • 1
    \$\begingroup\$ When Usain Bolt does the 100 metres in under 10s, then at t=0 his speed is also 0. Then after a small amount of time his speed is no longer 0... \$\endgroup\$
    – Solar Mike
    Jan 10, 2020 at 16:02
  • 2
    \$\begingroup\$ Your formula and Solar Mike both are correct. If current in the beginning is zero "my engineer sense says otherwise almost cant happen" then formula simplyfies. But your function probably gives forced response initial time. So your calculations also correct. \$\endgroup\$
    – emre iris
    Jan 10, 2020 at 16:07

3 Answers 3


I think i'm misinterpreting the meaning of this equation:

$$E(t) = \frac{L}{2} \bigg[i^2(t) - i^2(t_0)\bigg]$$

This is the amount of energy stored between time t_0 and t.

If I want the total energy stored in the inductor at time t...




and the equation becomes:

$$E(t) = \frac{1}{2} L i^2(t)$$


You cannot substitute an exponential equation for i(t) in "fixed time" equation for E unless E is changed to E(t) with a range of t and initial condition for current to know the value at any time (t).

Thus false calculations result which you have seen.


Your derivation gives the change of stored energy between t0 and t. But the "final" value of stored energy is still \$W = L \cdot i_{pk}^2/2\$ since the inductor has no energy stored in its magnetic field (because the initial current is zero since no voltage is applied) when before the voltage is applied. Once the voltage has been applied the current starts to ramp up from zero to its peak value according to \$V = L\cdot di/dt\$.

  • \$\begingroup\$ Can you clarify what you mean by "when the voltage is applied"? Clearly, the amount of stored energy is independent of the inductor voltage. \$\endgroup\$ Jan 10, 2020 at 16:11
  • \$\begingroup\$ Yes, the current is the time integral of voltage and reaches a max at I=ΔV/DCR where ΔV is the voltage across L \$\endgroup\$ Jan 10, 2020 at 16:49
  • \$\begingroup\$ @ElliotAlderson When no voltage is applied the current flowing through the inductor is zero. Thus, when the voltage is applied the current ramps up from zero. That's what I meant. Maybe I should have said "before" instead of "when". \$\endgroup\$ Jan 10, 2020 at 18:04
  • \$\begingroup\$ I think it would be clearer to say "when the voltage is first applied", but your change is also clear. \$\endgroup\$ Jan 10, 2020 at 20:11

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