The other day in class, our professor told us that care should be taken while closing a capacitive circuit and while opening an inductive circuit.

Why is this so?

I partly understand the inductive part. Since inductor only allows current to change gradually through it, if the current flow is suddenly stopped, then the voltage through it would suddenly increase, damaging it.

I'm unable to think of something for a capacitor.

  • 2
    \$\begingroup\$ "Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually? \$\endgroup\$
    – glen_geek
    Aug 18, 2019 at 15:58
  • \$\begingroup\$ @glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then? \$\endgroup\$
    – user_9
    Aug 18, 2019 at 16:49
  • 1
    \$\begingroup\$ Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing. \$\endgroup\$
    – glen_geek
    Aug 18, 2019 at 18:38
  • 2
    \$\begingroup\$ The capacitor itself probably has the least issue with a sudden short, that's more or less what it is built for, but all else might be highly problematic, like PCB traces starting to glow and ruin your day and such. \$\endgroup\$
    – wiebel
    Aug 19, 2019 at 9:37
  • 1
    \$\begingroup\$ Try it. It's instructive. \$\endgroup\$ Aug 19, 2019 at 9:48

3 Answers 3


In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.

  • \$\begingroup\$ But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit. \$\endgroup\$
    – Uwe
    Aug 18, 2019 at 16:28
  • 2
    \$\begingroup\$ Also, just for engineering concerns, even if the system can handle the inrush currents, the physical switch or relay itself must be robust enough to survive the associated arcing at the contact points, neither accumulating damage which could foul the contact nor welding itself closed. \$\endgroup\$
    – J...
    Aug 19, 2019 at 13:37
  • 1
    \$\begingroup\$ @J... Not to mention, accumulating damage that could present a physical safety hazard to the operator. \$\endgroup\$
    – bta
    Aug 19, 2019 at 16:25
  • \$\begingroup\$ That is an IDEAL capacitor, of course. \$\endgroup\$ Aug 30, 2019 at 19:00

edit /correction. ...

A step voltage cause a step current max Ic (t)=Vcc/ESR.
(@ t=0) when contact is made from discharged cap to an ideal voltage source then ....

Current decays at a slope by ~60% of initial surge max Ic at t=0 when t = ESR*C =\$\tau\$

So consider an expensive big cap with Vcc =12V, and ultra low ESR was 12 milliohm that in theory Ic max= Vcc/ESR=12V/12mohm = 1000 Amps

This is what happens If Vcc is a car battery is used as a capacitor of a value of many tens of thousands of Farads when the cold or CCA rating is only say 800 A.

But the same peak current occurs to a big plastic cap known to have low ESR but lower capacitance say with the same ESR. But when ESR is not given they rate the RMS ripple current or else it is called a general purpose meaning ESR*C is generally >> 100 us .

That might be an extreme example but illustrates the math .. otherwise the source Vcc would drop as C & Vcc charge up.

Computer grade caps or high voltage plastic caps conduct dangerous currents like boosting batteries with jumper cable. Reverse polarity charges can cause rapid breakdown and explosive results.


a Capacitor behaves short circuit in ideal when t=0 (initially) it means that if physical properties of such capacitor was suitable then the current flowing through cap would be infinite, but because of ESR it is not possible for real caps and for such ESR value initial current will be Vcap/ESR which is a very high amount of Amps For high amount capacities, Inrush current will be much more than a cap on a little circuit As you know high amount of Amps leads ARC when you apply voltage on a circuit, so it can damage if you behave recklessly

These conditions are the same for Inductive when you break the energized Inductor but the voltage will be a very high amount of Volts instead of current


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