Why does the current increase from zero to its steady state value in a circuit with a battery and an ohmic resistor, as opposed to initially reaching a very high value and then decaying down to the same value?

Mathematically, I can think of two possible reasons:

  1. Resistance is even larger initially (This seems unlikely considering the fact that the surface charges distribute instantaneously).
  2. It takes much more time to accelerate the electrons than it takes for the electric field to be established around the circuit.

3 Answers 3


When the switch is closed the electric field produced by the battery, and the force it applies to the electrons, appear almost instantaneously throughout the resistive circuit (near the speed of light). For a ideally purely resistive circuit there is no delay or overshoot in the current.

Only in circuits having inductance and/or capacitance, can there be some transient delay or overshoot in current, depending on where they are located in the circuit. For example, an ideal inductor in series with a resistor $R$ and ideal battery of voltage $V$, will delay the current from reaching its maximum of $V/R$. For an ideal capacitor in series with the resistor the current will initially be a maximum of $V/R$ and decay to zero.

But if current didn't have a transient state with a circuit with only a resistor, then how can there be a transient state in a circuit with an inductor? For an inductor to establish a potential there needs to be a changing current through the circuit.

The current is changing in the sense that before the switch is closed it is zero and after it is closed it increases from zero to the maximum due to the inductance which resists a change in current.

An inductor resists a change in current by producing an emf (potential difference) that opposes a change in current. Initially that emf is equal and opposite to the battery voltage for a net voltage across the resistor of zero and initial of zero. The induced emf across the inductor gradually decays so the voltage across and current through the resistor gradually (exponentially) increases to the maximum value of $V/R$.

For a series RL circuit the transient current is characterized by the following equation where $L$ is the inductance and $t$ is time:


and the transient voltage across the resistor is


Hope this helps.

  • $\begingroup$ But if current didn't have a transient state with a circuit with only a resistor, then how can there be a transient state in a circuit with an inductor? For an inductor to establish a potential there needs to be a changing current through the circuit. $\endgroup$
    – Piksiki
    Aug 5, 2022 at 20:13
  • $\begingroup$ @Piksiki I have updated my answer to respond to your follow up question. $\endgroup$
    – Bob D
    Aug 5, 2022 at 20:32
  • $\begingroup$ Just to make sure I understand what you mean correctly: since before the switch was closed the current was 0, immediately when the switch is closed the inductor will oppose any change in current? $\endgroup$
    – Piksiki
    Aug 5, 2022 at 22:05
  • 1
    $\begingroup$ Yes. The rule to remember is you can't change the current in an ideal inductor instantaneously (i.e., in zero time). Before the switch is closed the current in the inductor is zero since you have an open circuit. Since you can't change the inductor current instantaneously it has to also be zero the instant after the switch is closed at time t=0 in the equation I gave you. Thereafter the current builds to the max of V/R at time t = infinity. $\endgroup$
    – Bob D
    Aug 5, 2022 at 22:11

For the transient current to overshoot its steady-state value mathematically requires the presence of capacitance or inductance (or both) in the circuit. With only an (ideal, ohmic) resistor in it, overshoot is not possible.


The answers of both @Bob D and @niels nielson are perfectly correct within the framework of circuit theory, where the current is the same everywhere in the circuit, whether time-varying or not. However I think that @Piksiki is asking a different question: how does the state described by circuit theory get established, and what happens while that is happening?

I think that is a much more interesting question (hence my upvote) and much more difficult. I believe that a general answer backed by calculation is not known, because the answer depends on the geometry of the wires carrying the current from the source to the resistor. However there is one important case where we can give a detailed account of the few nanoseconds it takes for the steady state to get established, and that is for the particular geometry of a parallel conductor transmission line, such as a co-axial cable. In this case the current and voltage in the wire vary in space and time, satisfying the wave equation with a wave speed $v$ that depends on the construction of the cable but is usually a large fraction of the speed of light $c$.

So we have a voltage source $V_s(t)$ connected to one end of a transmission line of length $L$ with characteristic impedance $Z$, and a resistor $R$ connected across the other end. The general solution to the wave equation for the voltage and current at position $x$ and time $t$ is $V=f(t-x/v)+g(t+x/v)$ and $I=(1/Z)[(f(t-x/v)-g(t+x/v)]$, a sum of forwards and backwards-going waves. The arbitrary functions $f$ and $g$ are defined by the boundary conditions.

At the resistor end, $x=L$, the wave reflected back to the source is $r$ times the incoming wave where $r$, the reflection coefficient, is given by $(R-Z)/(R+Z)$. Thus, defining $T=L/v$, the time for a wave to travel from one end to the other of the transmission line, $g(t+T)=rf(t-T)$.

At the source end, the voltages must add up to $V_s(t)$: $f(t)+g(t)=V_s(t)$, or, substituting for $g$, $f(t)+rf(t-2T)=V_s(t)$. The solution of this is $$f(t)=\sum_{n=0}(-r)^nV_s(t-2nT).$$ In the case of $V_s(t)$ equal to a constant voltage $V_0$, switched on at $t=0$, this simplifies to $$ f(t)=V_0[1-(-r)^{p+1}]/(1+r)$$ where $p$ is the integer part of $t/2T$. This describes a voltage step propagating up and down the line, getting smaller on each pass and settling down to a steady value. The steady value is $f=V_0/(1+r)$ and $g=rV_0/(1+r)$. These describe a state with voltage $V=f+g=V_0$, as required, and current $I=(f-g)/Z=V_0/R$, as required. However they approach these final values either monotonically from below, if $r<0$, or in an oscillatory fashion from an initial overshoot if $r>0$. This directly answers the question: the current starts off at a value determined by the local configuration of the wires, and settles down at the correct value after a sufficient time for the travelling step to decay to a small value.

I suppose that something similar must occur with an arbitrary configuration of the wires, but I can't prove that. This is a long answer, but I think this is a difficult but important topic not discussed in textbooks.


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