# very simple question, plz I need the answer

Discussion in 'General Electronics Chat' started by fankoushat, Mar 28, 2010.

1. ### fankoushat Thread Starter New Member

Mar 8, 2010
17
0
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: Mar 28, 2010
2. ### dsp_redux Active Member

Apr 11, 2009
182
5
I don't see any circuits...?

3. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
There is no current in the circuit, so there is no drop across the resistor.

4. ### dsp_redux Active Member

Apr 11, 2009
182
5
$V=R*I$
If $I=0A$ from the open circuit, $V=0V$. Same thing beenthere said.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
1,123
But voltage between the open circuit terminals is equal Vsource

6. ### fankoushat Thread Starter New Member

Mar 8, 2010
17
0
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.

7. ### dsp_redux Active Member

Apr 11, 2009
182
5
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.

8. ### redlight000 Member

Feb 26, 2010
66
2
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??