Capacitance and Electrical Energy, Weird Question

Click For Summary
SUMMARY

The discussion centers on the challenge of using five 100 micro-F capacitors, each rated for a maximum voltage of 50V, to create a configuration that can handle a potential difference of 90V. The technician successfully combines two series capacitors in parallel with another two series capacitors, achieving a total capacitance of 100 micro-F. The maximum voltage across each capacitor in the parallel branches is confirmed to be 90V, as they share the voltage equally, thus meeting the requirement without exceeding the individual capacitor ratings.

PREREQUISITES
  • Understanding of capacitor configurations: series and parallel
  • Knowledge of capacitance calculations: C_{p} and C_{s} formulas
  • Familiarity with voltage ratings and potential differences in electrical circuits
  • Basic electronics terminology, including microfarads (μF) and volts (V)
NEXT STEPS
  • Study capacitor series and parallel configurations in detail
  • Learn about voltage division in series circuits and its implications
  • Explore practical applications of capacitors in power supply circuits
  • Investigate safety ratings and derating factors for capacitors in high-voltage applications
USEFUL FOR

Electronics technicians, electrical engineering students, and hobbyists working on power supply repairs or capacitor configurations will benefit from this discussion.

AznBoi
Messages
470
Reaction score
0

Homework Statement


To repair a power supply for a stereo amplifier, an electronics technician needs a 100 micro-F capacitor capable of withstanding a potential difference of 90V between its plates. The only available supply is a box of five 100 micro-F capacitors, each having a maximum voltage capability of 50 V. Can the technician substitue a combination of these capacitors that has the proper electrical characteristics, and if so, what will be the maximum voltage across any of the capacitors used? [Hint: The technician may not have to use all the capacitors in the box}


Homework Equations


C_{p}=\Sigma C_{i}
\frac{1}{C_{s}}=\Sigma \frac{1}{C_{i}}


The Attempt at a Solution


I just fiddled around with the combinations until I arose to one with a perfect total capacitance of 100 micro-F.

My combination: 2 series capaciators that are parallelly linked by another 2 series capacitors.

2 series capcitors= 50F 2 parallel 50F capacitors= 50F+50F=100F

I also attempted to find the maximum voltage across the capacitors and I came up with 100V for each of the 2 parallel capacitors. Since, in parallel circuits, the velocity is constant, does that mean the maximum voltage across any of the capcitors used is 100V??

If my description of my combination befuddles you, I will be glad to draw a diagram to go along with it. :smile: Thanks for your help in advance.
 
Physics news on Phys.org
AznBoi said:
My combination: 2 series capaciators that are parallelly linked by another 2 series capacitors.

2 series capcitors= 50F 2 parallel 50F capacitors= 50F+50F=100F

Yes, that works.

I also attempted to find the maximum voltage across the capacitors and I came up with 100V for each of the 2 parallel capacitors.

How? You're only applying 90V, right? So that's the potential difference across each parallel branch. And since the two capacitors in each branch are identical, it should be straightforward to show that they share that 90V equally.

Since, in parallel circuits, the velocity is constant, does that mean the maximum voltage across any of the capcitors used is 100V??

velocity? :confused:
 
Tom Mattson said:
Yes, that works.



How? You're only applying 90V, right? So that's the potential difference across each parallel branch. And since the two capacitors in each branch are identical, it should be straightforward to show that they share that 90V equally.



velocity? :confused:

V is volts. Sorry, I know that the letterings and symbols are soo confusing. Anyways, I don't get the voltage part. It says: capacitor capable of withstanding a potential difference of 90V between its plates. I suppose that means that the capacitators have to have a minimum of 90V?? I found the 100V on each branch by adding the series voltage up so 50V+50V=100V per branch. Since 100V and 100V are parallel, doesn't that make the whole system 100V??

I just don't undestand if they already give you a power supply. Is the battery 90V or what?
 
No, I don't think you've read the problem right. If the repairman needs a single equivalent capacitance that can handle 90V, then I think you are to assume that he intends to put 90V across it. So you've got something like this:

Code:
        -----||-----||-----
        |                 |
        |                 |
        -----||-----||-----
        |                 |
        |------[90V]------|
 
Tom Mattson said:
No, I don't think you've read the problem right. If the repairman needs a single equivalent capacitance that can handle 90V, then I think you are to assume that he intends to put 90V across it. So you've got something like this:

Code:
        -----||-----||-----
        |                 |
        |                 |
        -----||-----||-----
        |                 |
        |------[90V]------|

Yep, that's exactly what the combination looks like! However, does that mean you just disregard the part: "each having a maximum voltage capability of 50 V??" Does that piece of information mean anything for part b)? If not, I guess the maximum voltage would be 90V??
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 67 ·
3
Replies
67
Views
12K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
10
Views
5K