# very simple question, plz I need the answer

#### fankoushat

Joined Mar 8, 2010
17
Hi guys,

Consider the following circuit:

http://www.postimage.org/image.php?v=gxFeffA

V source , connected in series with resistor R , and open circuit.

what is the voltage dropped between the open circuit terminals?

equal to V source, or equal to V source - V dropped on R?

Thanks alot

Last edited:

#### dsp_redux

Joined Apr 11, 2009
182

#### beenthere

Joined Apr 20, 2004
15,819
There is no current in the circuit, so there is no drop across the resistor.

#### dsp_redux

Joined Apr 11, 2009
182
$$V=R*I$$
If $$I=0A$$ from the open circuit, $$V=0V$$. Same thing beenthere said.

#### Jony130

Joined Feb 17, 2009
5,163
But voltage between the open circuit terminals is equal Vsource

#### 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?? 