I am curious about how the time bounds in power (t>=0+) and energy (t>=0) are developed in the natural response of RL and RC circuits. For the RL circuit, the natural response of the current is $$i(t)=I_0e^{-(R/L)t}=I_0e^{-t/\tau}, \, t\ge0$$ because there can be no instantaneous change in current to an inductor acting as a short-circuit when the inductor experiences a constant current for a long time. Because there is no voltage across the inductor initially, I see that the voltage is only bounded to time after the switch is opened since the voltage exactly at t=0 is indeterminate, $$v(t)=I_0Re^{-t/\tau}, \, t\ge0^+.$$ I am wondering why the power dissipated to resistive components then is bounded as $$p=I_0^2Re^{-2t/\tau}, t\ge0^+$$ whereas the energy delivered to resistive components is bounded as $$w=\int_0^\infty p\,dt = \int_0^\infty I_0^2 Re^{-2t/\tau}dt, \,t\ge0$$ in the basic circuit example where the inductor is suddenly removed from a current source and undergoes the natural response decay. In some way, I can see that the power should be bounded to the t=0+ side, if the voltage is not defined at t=0, however the power should be able to be derived with any of the equations, $$p=vi=v^2/R=i^2R$$ where I would think that I should be able to find the power with the last expression even at strictly t=0. Furthermore, I am most curious to know why the energy can be integrated from t>=0, if the power is only defined from t>=0+. If an definite integral gives the instantaneous rate of change between two bounds, I thought that the range of t should be only defined for t>=0+.

edit update: I think i(t) would be the current going through the resistor R branch, in the time interval t>=0 based on the picture that the text showed in Figure 7.3. The initial current in the circuit is the source current short-circuited through the inductor branch and is explained as, $$I_s=i(0^-)=i(0^+)=I_0, \, t<0$$ which is the current running through inductor that was acting like a short circuit right until the current source is removed, and so in the resistor branch, $$i(t)=i(0)e^{-t/\tau}=I_0e^{-t/\tau}, \, t\ge0$$ $$i(t)=0, \, t<0$$.

An RL circuit

better arranged equations in question

  • \$\begingroup\$ there can be no instantaneous change in current to an inductor under any circumstances. Please explain what i(t) and \$I_0\$ are representing. \$\endgroup\$
    – Andy aka
    Commented Oct 28, 2023 at 20:03

1 Answer 1


If we assume that the writer of Fig 7.3 is serious, he has actually said that the switch has stopped conducting when the time is exactly =0. There's no last moment of time when the switch conducted, but the first non-conducting moment is t=0. That's said in the validity range of formula i(t). That may be cumbersome, but it really is printed there. And in math it's perfectly valid.

If the last conductive state of the switch occurred at t=0 the validity range of the given formulas should be t>0.

But the writer has lost his confidence and taken a half step back. He has introduced something really obscure: the number 0+. I guess he has in that phase thought the positive t which is closest the zero. That's nonsense, such number is non-existent in math. It could exist in some digital system where time is measured as finite steps.

The formulas are valid when 0+ is changed to 0.

  • \$\begingroup\$ Do you mean for i(t), v(t), p(t), w(t) for all being in the t>0 range and not t>=0? \$\endgroup\$
    – shinyleaf
    Commented Oct 29, 2023 at 4:24
  • 1
    \$\begingroup\$ @shinyleaf i(t) and w(t) are already OK. remove 0+ from the validity range of v(t) and p(t). Let their validity ranges also be "t greater than or equal with zero". The 0+ is meaningless. The first line (that's i(t)) has nailed that the switch is in off-state at t=0 and after it. \$\endgroup\$
    – oneprivate
    Commented Oct 29, 2023 at 5:34

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