Consider the following circuit. I'm tying together the basic definitions of capacitor and inductors, and how they work when connected to a DC source.

Circuit displaying a steady-state RLC circuit

First off, we have the equations for current throuh a capacitor, and voltage across a inductor.

  • \$i_c(t) = C\frac{dv_c(t)}{dt},\quad v_l(t) = L\frac{di_l(t)}{dt}\$

Analyzing this, we can see, clearly, that if our inductor and capacitor are "empty" at \$t=0\$, that our capacitor acts as a short circuit, as there's no current going through it unless there's a change in voltage across the capacitor.

However, I'm having difficulties seeing how the inductor acts as a short circuit. Because we have the DC voltage source \$10V\$, all we know is that our voltage is constant. What I wonder then is how we can make the statement that our inductor acts as a short circuit.

Also, why do we call direct voltage sources "DC voltage source"? Doesn't is sound a bit strange saying "direct current voltage source"?

  • 1
    \$\begingroup\$ You've asked two unrelated questions. I'd suggest splitting out the one about naming of DC into a separate question. \$\endgroup\$ Sep 18, 2015 at 19:21

2 Answers 2


For capacitor ic(t)=C *dvc(t)/dt, when switch is closed there is voltage difference between the +ve and -ve terminal. Hence, there will be flow of current until cap is charged. Since battery is a constant voltage source, the cap will be charged to 10 V and after that there won't be any flow of DC current through it and it will act as open circuit for DC current.

In case of inductor, vl(t)=L *dil(t)/dt, vl(t) is the voltage across the inductor, hence when circuit is closed there is huge di/dt in transisent state and inductor will act as huge resistor. But as the current becomes constant at steady state, di/dt = 0, V(l) = 0 which means voltage across inductor is zero hence short.

The reason for calling the source as voltage source is that it provides constant voltage for different loads which may draw different current.


I think perhaps you're confusing steady-state analysis and transient analysis. Steady-state analysis doesn't concern itself with what happens at the moment the switch closes. It assumes the switch has always been closed, and that the system has settled into a static (or permanently oscillating) state.

If you close the switch and wait long enough, the current and voltage will stop changing; there's nothing to keep them changing, after all, it being a DC system.

Since dv/dt is now zero, current through the capacitor is zero, but there's still voltage across it; that's an open circuit. Since di/dt is now zero, voltage across the inductor is zero, but there's still current flowing through it; that's a short circuit.


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