very simple question, plz I need the answer

Thread Starter


Joined Mar 8, 2010
thanks johny, but why people say " the inductor at the beginning of connecting it with a dc source, will have voltage difference equals to V source - V dropped on R, because it is considered as open circuit".

So voltage difference measured between the two open circuit terminal will be the same as V source.


Joined Apr 11, 2009
When you do a DC analysis, you can consider capacitors as "open circuit" and inductors as "short circuit". There is no voltage drop in a short circuit. DC voltage is considered as a "0 frequency" or 0Hz. Impedance of a capacitor is \(X_C = \frac{1}{2\pi f C}\). If you take \(f\rightarrow 0\), \(X_C=\infty\) hense the capacitor considered as an open circuit. For the inductor \(X_L = 2\pi f L\). \(X_L = 0\) when \(f=0\). It is considered as a short circuit.


Joined Feb 26, 2010
Hi dsp,
Just asking how do you get those electronic equations on the post? are they built in this actual forum?? or you got a special program to do it??

Why I ask I've got a few maths questions to ask..
many thanks