Electronics Circuit question regarding capacitors.

solidMGSsnake

What happens if the microFarad value of a capacitor is changed in a Series + Parallel Circuit that contains LEDs?

This keeps coming up in my electronic exams yet we have no notes in it that the lecturer gave us. In fact my lecturer gave me little notes, I have very little understanding of electronics. Also googling did nothing. I have a an exam in a few days yet I can't find answer anywhere.

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.

I asked this question a few days ago on yahoo answers but the only answer I got was:

"if 1000 uF capacitor voltage is V and you connect a 22 uF to it...
(1/2 ) 1000 x V^2 = (1/2 ) 1022 V' ^2 so V' = V x 0.98"

I understand very little of that. I'm not adding a capacitor to it, I'm swapping them out. Is it possible to explain it in plain english? As this doesn't help at all. I tried asking him what it meant but I got no answer.

Please try help me.

brightspark

Can you draw the circuit here?

Short answer is it may affect current or voltage?

Do the calculations on the circuit with both values of capacitor, see what changes!

solidMGSsnake

What calculations? This is what I don't know what to do. How can I draw the circuit.

TheChizler

I'm guessing it's sort of a trick question, presuming it's a DC circuit because of the LEDs. What happens to capacitors in a DC circuit? How would it affect an LED in parallel to the cap?

solidMGSsnake

I don't know, that's why I'm posting this question. I know next to nothing on electronics. My lecturer was this old guy who mumbled and never gave us any notes and sometimes didn't come to class.

brightspark

solidMGSsnake wrote: »

What calculations? This is what I don't know what to do. How can I draw the circuit.

Then take a photo of the circuit in the question and post it?

solidMGSsnake

I can't I don't have a digital camera and I lost the cable belonging to my phone. xD It doesn't have to matter how the circuit actually is. In fact forget that particular circuit. Answer the question for any simple circuit you want, I can apply it to mine afterwards. All I nned to know is how to do it.

brightspark

http://lushprojects.com/absolutebeginners/elecabsbegin.pdf

Is it similar to page 9?

You should probably study it all.

solidMGSsnake

No that's like using 2 circuits at the same time. I'm not moving any components. In a parallel circuit with LEDs and Button (they have to matter, just a parallel circuit) what would happen if the capacitor was swapped from a 22mf to a 100mf. Not while the circuit is running. Imagine of two of the same circuits, expect the capacitor is 22mf in one and 100mf in the other, does it behave differently? For example, is the LED brighter, is voltage lower?

king_of_inismac

Is this a dc or ac circuits. That does matter.

Where are the capacitor located in relation to the LEDs and buttons. Again that does matter.

The reason people are asking for more details (like a circuit diagram) is that the answer depends on the above.

TheChizler

Do you have any books on circuit analysis or any notes at all? Someone must have taken notes in class. Even Wikipedia would help.

In a water analogy, imagine a capacitor as a rubber diaphragm blocking a pipe, it will fill up and expand with pressure and return to it's original position once pressure decreases. It won't let any net flow through. IE DC current won't flow through a capacitor, once it's charged it acts as an open circuit.

Can I ask what college you're in? The paper you're doing might be available online.

ZENER

I'd have a guess that this is to do with the charging and discharging of the cap through the LED. One switch is in series with the supply and the CAP to charge it, the LED and the second switch are in series across the CAP or in parallel if you like.

The first switch is pressed to charge the 22uF capacitor then released. The second switch connects the LED to the now charged CAP causing it to discharge through the LED which will light for a period of time. Replacing the 22uF CAP with the 1000uF CAP will cause the LED to light for longer.

Possible ?

Ken

TheChizler

The diagram would really help OP. You could draw one using ASCII characters, like here, or alternatively use Microsoft paint, or just describe in words each node of the circuit.

ZENER

A bit old tech but all I have to hand at the moment. Hopefully it's intelligible ?



Ken

TheChizler

Ah I failed to give any consideration to the switches, most likely not a trick question afterall.

solidMGSsnake

I seriously have no notes, I'm telling you the lecturer is useless. I'll see if I can get a copy of the exam or at least one that has a similar question, I'll post some kind of circuit for ye later.

It doesn't say but I assume it's DC since the battery is made from 2 x AA Batteries. The exam has the circuit as a breadboard, but I was never good at reading them. If I drew up the breadboard on here, would that be ok? Also if ye could convert the breadboard into a schematic diagram that would help me as well.

I believe the order from my limited knowledge of reading breadboards is Button1 -> Resistor1 -> LED1 -> LED2 -> Resistor2 (-> Button2 or
-> Capacitor (other direction))

That probably didn't help and is probably wrong, I'll try up the breadboard if ye like.

ZENER

Go for it !!

riffmongous

ZENER wrote: »

I'd have a guess that this is to do with the charging and discharging of the cap through the LED. One switch is in series with the supply and the CAP to charge it, the LED and the second switch are in series across the CAP or in parallel if you like.

The first switch is pressed to charge the 22uF capacitor then released. The second switch connects the LED to the now charged CAP causing it to discharge through the LED which will light for a period of time. Replacing the 22uF CAP with the 1000uF CAP will cause the LED to light for longer.

Possible ?

Ken

With out any more information I'd guess this is probably going to be the answer

solidMGSsnake

riffmongous wrote: »

With out any more information I'd guess this is probably going to be the answer

Ya I agree that could be it, we'll see after I draw up the breadboard.

I won't be able to for a few hours though as I don't have the paper with me.

TheChizler

Does anyone else have notes you could borrow?

solidMGSsnake wrote: »

The exam has the circuit as a breadboard, but I was never good at reading them. If I drew up the breadboard on here, would that be ok? Also if ye could convert the breadboard into a schematic diagram that would help me as well.

Always redraw if it helps understanding!

solidMGSsnake

Here's question 1

https://scontent-b-vie.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/1621717_753974294645018_5810237061013094868_n.jpg?oh=86a213a7cacdbdf83b9a8a30f4653f02&oe=546F11FD

ZENER

Hmm . . . doesn't really make much sense. Can't really see the point of the circuit. But this is what I see:



In the questions it says to assume that there is a volt drop of 0.6V across the LED, LEDs drop 2.0V typically so that's what I don't get. Pressing the right button should cause both LEDs to dim momentarily (providing the Left button is held on) while the CAP charges, probably barely noticeable with the 22uF but a bit more so with the 1000uF (depends on the internal resistance of the battery). The currents can be calculated using Ohms Law. I'm assuming this is a college assignment so I'm reluctant to answer the questions for you.

Ken

solidMGSsnake

A friend just sent me this:



" the operation of the circuit is that when you push the first push button the led at r1 will light up, if you push both push buttons the led at r1 and r2 will light up. then after if you release push button 2 led at r2 will dim over time as there as a capacitor on the same connection as the led 2."

ZENER

solidMGSsnake wrote: »

A friend just sent me this:



" the operation of the circuit is that when you push the first push button the led at r1 will light up, if you push both push buttons the led at r1 and r2 will light up. then after if you release push button 2 led at r2 will dim over time as there as a capacitor on the same connection as the led 2."

The value are wrong for the resistors and the battery but the description matches. If you press both buttons then both LEDs will light up, releasing the buttons will cause the left LED to extinguish but the right one will stay lit for T=RxC (in seconds). The battery is 3 volts DC and the two resistors look like Yellow Purple Brown which is 470 ohms.

Ken

solidMGSsnake

ZENER wrote: »

The value are wrong for the resistors and the battery but the description matches. If you press both buttons then both LEDs will light up, releasing the buttons will cause the left LED to extinguish but the right one will stay lit for T=RxC (in seconds). The battery is 3 volts DC and the two resistors look like Yellow Purple Brown which is 470 ohms.

Ken

Ok thanks, but that doesn't answer my question regarding the capacitors, or if it does, i don't get it.

solidMGSsnake

Also what's wrong with the value of the battery?

I don't think you are faulted with the resistors being wrong. The breadboard diagram doesn't give their value besides the color code which I don't think they expect you to remember all of them. Their value was only mentioned further down in the question, part (iv).

ZENER

solidMGSsnake wrote: »

Also what's wrong with the value of the battery?

I don't think you are faulted with the resistors being wrong. The breadboard diagram doesn't give their value besides the color code which I don't think they expect you to remember all of them. Their value was only mentioned further down in the question, part (iv).

The breadboard shows two AA cells at 1.5V each which is 3V in total as they are in series. The questions says that the resistors are 470 ohms.

If you are serious about electronics then you WILL need to memorise the colour codes as they apply not only to resistors but capacitors and even diodes !

Completely politically incorrect (sincere apologies but not in the early 80's though !?) but the rhyme was as follows (starting with 0 ) Big (Black) Boys (Brown) Rape (Red) Our (Orange) Young (Yellow) Girls (Green) But (Blue) Violet (Violet or Purple) Gave (Grey) Willingly (White) represented 0 1 2 3 4 5 6 7 8 9.

Hopefully I haven't offended anyone but that was the rhyme I was taught.

Ken

solidMGSsnake

Wow that rhyme is awful but helpful xD
No seriously I don't think we actually have to remember the color codes. It was never looked at in detail during the course. While it says it's 470, that was later down so if you're doing the exam part by part, wouldn't have seen that yet, so I don't think they would fault you.

Pehaps you're right about the 3V battery, but it might have just been a representation.

Anyhow can you try answer the question in the OP? I really have no idea.

ZENER

The voltage is rather important. Of course it depends on the course you are doing. What is it by the way ?

Ken

TheChizler

Bit of a weird circuit, but what Ken said, the time the led stays on for will correspond to R x C.

solidMGSsnake

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.