Electronics Circuit question regarding capacitors.

Xantia

All knowledge in this area is helpful even if you cannot see that right now.
For example some capacitors in power supplies dry out, they will work away for years but if you turn the power off and on again the starting capacitors will not load up the power supply.

solidMGSsnake

So no one has a concrete answer regarding the capacitor change?

Bigdavyj

The answer to your first question is: When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors. An increase in plate area, with all other factors unchanged, results in increased capacitance.
When capacitors are connected in series, the total capacitance is less than any one of the series capacitors' individual capacitances. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. An increase in plate spacing, with all other factors unchanged, results in decreased capacitance.

If you've built your own computer as you told us earlier this stuff should be simple to u

Chet T16

http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0As+304+144+464+144+0+1+true%0Ar+304+224+304+144+0+470.0%0Ar+464+224+464+144+0+470.0%0A162+304+224+304+304+1+0.6+1.0+0.0+0.0%0A162+464+224+464+304+1+0.6+1.0+0.0+0.0%0Aw+464+144+544+144+0%0Aw+464+304+544+304+0%0Aw+160+304+304+304+0%0Aw+304+304+464+304+0%0As+160+144+304+144+0+0+false%0Ac+544+304+544+224+0+2.2E-5+-0.36368472950619224%0Ar+544+144+544+224+0+10.0%0Av+160+304+160+224+0+0+40.0+1.5+0.0+0.0+0.5%0Av+160+224+160+144+0+0+40.0+1.5+0.0+0.0+0.5%0A

Johnboy1951

solidMGSsnake wrote:

Here's the exact question: "What would the effect be if the 22 microfarad polarised capacitor was changed to a 1000 microfarad capacitor? Briefly explain." The circuit in question is a parallel circuit with 2 buttons and 2 LEDs. If you could also tell me if the effect would be different in a series circuit just in case. Thanks.

It depends on the circuit.

Based on the circuit you posted from a friend, increasing the capacitor value will increase the time the LED will stay illuminated after the battery has been disconnected (as the charge on the capacitor discharges through the resistor and LED).

ZENER

Didn't I already say this ?! T=RxC ! No offence OP but you can't expect us to do all the work for you, this a course for learning you're doing - so learn !

Ken

Johnboy1951

ZENER wrote: »

Didn't I already say this ?! T=RxC ! No offence OP but you can't expect us to do all the work for you, this a course for learning you're doing - so learn !

Ken

Yes, but clearly not in a form the O-P understood

ZENER

Johnboy1951 wrote: »

Yes, but clearly not in a form the O-P understood

Check Post 12 on page 1!

Anyway multiple posters have posted regarding this with enough info for most people to figure it out.

I don't believe in doing someone else's homework so in true Dragons Den parlance - I'm out !!

Ken

DublinDilbert

solidMGSsnake wrote: »

The course itself is game development but the module is Electronics for Computing which really has nothing to do with anything. Everyone at the college says it's useless for the course. I've built my own computer and I still didn't need to know any of this stuff. Game development for the most part is software.

What happens when you need to interface to a game controller or other electronic hardware?

When you say "everyone at the college" do you mean the other students who don't want to take the module?

solidMGSsnake

Sorry ya I meant everyone in my course, even the higher years said that they never liked it and it was worthless.

ZENER wrote: »

Didn't I already say this ?! T=RxC ! No offence OP but you can't expect us to do all the work for you, this a course for learning you're doing - so learn !

Ken

Chill out O_o
The whole reason I reason I posted this is because I don't have any notes and the internet didn't help. So don't tell me to just magically learn. No need to be mean. I'm worried enough about failing without getting a lecture from a guy I don't even know
I did read your post and I acknowledged it, but you only said it was a possibility, you never actually said it was correct, so I kept looking for a definite answer. If I knew it was correct, I would have ended the post there.

Bigdavyj wrote: »

The answer to your first question is: When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors. An increase in plate area, with all other factors unchanged, results in increased capacitance.
When capacitors are connected in series, the total capacitance is less than any one of the series capacitors' individual capacitances. If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. An increase in plate spacing, with all other factors unchanged, results in decreased capacitance.

If you've built your own computer as you told us earlier this stuff should be simple to u

I'm sorry but I don't understand much of that. To me it's like saying water has "waterance". I assume capacitance is the level of ability a capacitor has? I have never come across that term in my course. I don't know what a plate area is either. I'll try google those terms and I'll probably be able to follow what your saying. Or you can post the definitions if you don't mind.

Also have you built a PC before? Because no you do not need to know any of that for building one.

I appreciate all work guys and I'm only asking for help cause I'm desperate. If I'm annoying you I'm sorry, but I'm not forcing you to do anything. I have very limited knowledge of electronics. It's like I've never done it before because the classes didn't teach us anything! I only know what I thought myself via internet.

TheChizler

Capacitor is to capacitance as balloon is to size-of-the-balloon.

Even a look at wikipedia:capacitor would give you an idea of what's involved. You're aware that current is just how many charged particles (electrons) flowing around a circuit. A capacitor is device that can store that charge and release it in the opposite direction.

If you apply a voltage to a capacitor, current will flow in to it until it is 'full'. How full it will get depends on it's capacitance, which, similar to how resistance states how much current will flow through a resistor depending on a voltage (I = V/R), says how much charge will be stored in the capacitor, depending on the applied voltage.

Q = VxC, where Q is the charge, or number of electrons stored, and C the capacitance.

So the bigger the C the more charge can be squeezed in for the same voltage. Once the voltage applied to the capacitor has been removed, the capacitor will have it's own voltage, which basically will be the same as the voltage used to charge it.

A charged capacitor can be used similarly to a battery, to power things until its charge runs down. If you were to connect something to the capacitor, if its voltage is large enough, it would power it for a time.

The time a capacitor takes to charge (through a resistance, everything has a resistance or equivalent resistance) to 63.2% , or discharge to 36.8% is τ, or the RC time constant (τ = RxC). This is a standard used, as you can't say how long it takes a capacitor to fully dis/charge as technically this would take infinite time. Keeping everything else the same, this time of course depends on the value of C.

solidMGSsnake

Thank you so my answer would be to do the formula Q=VxC twice with subbing C with 22 first time and 1000 the second time, is that right?

TheChizler

solidMGSsnake wrote: »

Thank you so my answer would be to do the formula Q=VxC twice with subbing C with 22 first time and 1000 the second time, is that right?

You could that, and explain the effect this change has on the LED.

raspberrypi67

I can help you out here, I'm an electronic engineer. To give you some basis. In a DC circuit, Electrolytic capacitors or polarised caps, are like
reservoirs, if they are not charged up already and you apply a voltage from a DC power supply or battery, it'll charge pretty rapidly, if there is no resistance involved.
With PB1 pressed briefly, your cap will charge within a micro second!! ( this is assuming that PB2 is OPEN ).
Ok, now your cap is charged. Depending on how big your cap is, close PB2 and you may well see a brief flash of the LED as the cap discharges into it.
The larger the cap say going from 22uF to 1000uf you'd see a brighter flash. In terms of uF, the larger the cap value the more charge iot can hold.
My pc battery is running low, I'll post this, U can let me know if you need more info

R