# step-by-step

Discussion in 'Homework Help' started by Voltboy, Jan 10, 2007.

1. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
i was watching this and im confused about how this work.. what are the capacitors here for.. pllease someone explain me why you need each coponent in this circuit?

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Hi,

May I suggest you read Volume III of the e-book this site hosts?

3. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
im reading it, thats how im learning all but for me using a capacitor there dont make sense because the battery is always producing current , so the capacaitors will always be charging and acting a a load always, so they when cahrged will never be discharged and work like a simple conductor/wire, so it dont make sense

4. ### Distort10n Active Member

Dec 25, 2006
429
1
The coupling capacitors or the tuning circuit?

5. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
i dont know what is each of those

6. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
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can anyone tell me in what direction goes the current, i know where it ends but i cant find all the current path

7. ### mikeonthegreen Member

Jan 7, 2007
21
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OK I am new to electronics, but as I understand it a capacitor blocks dc and allows ac to pass.
Also I dont think caps act as a wire when charged do they?
Dont they charge and discharge so there would not be a steady flow of current. I am speculating but I hope someone can correct me.

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Pretty much. The "current flow" in a capacitor is due to the changing voltage across the capacitor: i(t) = C(dV/dt)

The voltage stops changing (i.e. dv/dt = 0), then the plates of the capacitor charge to a steady state condition and the "current" ceases to flow.

Dave

9. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
But, if the capacitor is charged and its still current flowing will the capacitor discharge or it will only discharge if the current stop flowing (if u disconnect the battery)

May 16, 2005
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11. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Thinking of a DC application, then when the capacitor is charged, the voltage across the capacitor is constant, i.e. dv/dt = 0 and therefore i = 0. The situation changes for AC application, because the opposing plates of the capacitor are constantly charging and discharging, i.e. dv/dt /= 0 and therefore i /= 0.

The crucial point to note is, that the voltage across the capacitor must be changing for current flow to occur. Without the need to complicate things, but dependant on the dielectric properties of the dielectric between the plates you may experience some current flow from one plate to another, although this is negligible in capacitor components, it has a noticable impact on dielectric heating processes which exploit the same mechanism as charging and discharging of a capacitor.

Dave

12. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
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oh thanks dave.. umm, last question, where is the negative end of the source in the circuit i showed in the post, the ground right?

13. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Yup. Most common convention is to show any voltages at reference to ground, so if only one voltage is shown, the opposite power connection will be ground.

14. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
Thanks alot to everybody that helped me