very simple question, plz I need the answer

Thread Starter

fankoushat

Joined Mar 8, 2010
17
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.
 

dsp_redux

Joined Apr 11, 2009
182
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.
 

redlight000

Joined Feb 26, 2010
66
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
from
redlight000
:D
 
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