Do capacitors act as an open circuits or closed circuits at time t=0? Why? What about inductors? I tried it out, and what I got was this: Initially when I opened the switch, the capacitor acted like a short circuit. That should not be happening, right? A capacitor should block DC. I tried with a couple different caps. I am very confused.
20.2k 8 8 gold badges 58 58 silver badges 102 102 bronze badges asked Dec 3, 2009 at 14:50 ruchi ruchi\$\begingroup\$ What about WHAT inductor? It would probably be best to provide details for the network in question. Also, if you have access to a lab I suggest trying it out. Seeing it really helps you grasp what's going on. \$\endgroup\$
Commented Dec 3, 2009 at 14:57\$\begingroup\$ A capacitor looks like an open circuit to a steady voltage but like a closed (or short) circuit to a change in voltage. And inductor looks like a closed circuit to a steady current, but like an open circuit to a change in current. \$\endgroup\$
Commented Nov 30, 2011 at 21:36\$\begingroup\$ You probably should put this as an answer, as I believe that is what the OP is looking for. Perhaps with a brief explantion as to why (cap charging and magnetic fields and such). \$\endgroup\$
Commented Dec 8, 2011 at 18:49\$\begingroup\$ @Tuva - Thanks, though I can't take all the credit - it was an improvement on a suggested edit. \$\endgroup\$
Commented Dec 8, 2011 at 21:15\$\begingroup\$ @ChrisStratton I think it would be much easier for the OP to understand if you speak of the characteristics of these circuit elements in terms of their impedance in different applications instead of memorizing what they 'should' be. Although, this post is old so he most likely got it. \$\endgroup\$
Commented Aug 2, 2013 at 12:20Short Answer:
Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor)
Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor)
Long Answer:
A capacitors charge is given by \$Vt=V(1-e^<(-t/RC)>)\$ where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.
At the exact instant power is applied, the capacitor has 0v of stored voltage and so consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.
For an inductor, the opposite is true, at the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), as time continues, it will have a low resistance to the steady voltage and carry lots of current (short circuit).