Series Combo Of Capacitors, Smaller Than Single Capacitor

  • Thread starter BOAS
  • Start date
  • #1
555
19

Homework Statement



"A series combination of capacitors must always be smaller than the smallest individual capacitor."

This is the statement that I am trying to validate, but I am stuck.

Homework Equations




The Attempt at a Solution


[/B]
I know that if you have a circuit that consists of a cell and a number of capacitors in parallel, the overall effect due to the 'inside' plates neutralising each other, is that of one capacitor with a large plate separation. This large plate separation is what causes the equivalent capacitance to be lower than the sum of it's parts, but I don't understand how I can show that this value must be smaller than that of the smallest capacitor in the circuit.

I can show that it is the case for specific examples, but not a general case.

Thanks for any help you can give,

Jacob.
 

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
Perhaps start by showing that 1/Ceq = 1/C1 + 1/C2 + 1/C3 etc
 
  • #3
gneill
Mentor
20,913
2,862
Have you obtained the formula for the equivalent capacitance of two capacitors in series? (You can always deal with more than two by combining two at a time).
 
  • #4
555
19
Perhaps start by showing that 1/Ceq = 1/C1 + 1/C2 + 1/C3 etc
Ok,

[itex]V = \frac{Q}{C}[/itex]

So the voltages on the individual capacitors is given by;

[itex]V_{1} = \frac{Q}{C_{1}}[/itex],

[itex]V_{2} = \frac{Q}{C_{2}}[/itex],

[itex]V_{3} = \frac{Q}{C_{3}}[/itex] etc...

The total voltage is [itex]V_{tot} = V_{1} + V_{2} + V_{3} + ...[/itex]

[itex]V = \frac{Q}{C_{eq}} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} + ...[/itex]

Dividing through by [itex]Q[/itex] yields

[itex]\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + ...[/itex]
 
  • #5
gneill
Mentor
20,913
2,862
Okay, now using what you've done, write and simplify the equation for the equivalent capacitance of just two capacitors in series.
 
  • #6
555
19
Okay, now using what you've done, write and simplify the equation for the equivalent capacitance of just two capacitors in series.
Okay,

[itex]\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}[/itex]

[itex]C_{1}C_{2} = C_{eq}C_{2} + C_{eq}C_{1}[/itex]

[itex]C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}[/itex]
 
  • #7
gneill
Mentor
20,913
2,862
Excellent. Now does that last expression suggest anything to you about showing that ##C_{eq} < C_1## and ##C_{eq} < C_2##?
 
  • #8
555
19
Excellent. Now does that last expression suggest anything to you about showing that ##C_{eq} < C_1## and ##C_{eq} < C_2##?
Not really...
 
  • #9
gneill
Mentor
20,913
2,862
What if I write the expression this way:
##C_{eq} = C_1 \frac{C_2}{C_2 + C_1}##
 
  • #10
555
19
What if I write the expression this way:
##C_{eq} = C_1 \frac{C_2}{C_2 + C_1}##
Ah ha!

##C_{eq} \propto C_{1}##

##C_{eq} = k C_{1}##

##k = \frac{C_{2}}{C_{2} + C_{1}}##

##C_{2} < C_{2} + C_{1}## (Capacitors can't have negative values)

Therefore ##k < 1##

and

##C_{eq} < C_{1}##

The same argument applies to ##C_{2}##

That makes a lot of sense to me, but is the argument general enough to apply to any number of capacitors?.
 
  • #11
gneill
Mentor
20,913
2,862
That makes a lot of sense to me, but is the argument general enough to apply to any number of capacitors?.
See what you can do with what you have now and the comment I made in post #3.
 
  • #12
249
26
What if I write the expression this way:
##C_{eq} = C_1 \frac{C_2}{C_2 + C_1}##
For that, wouldn't it be necessary that C1+C2>1 ?
 
  • #13
555
19
See what you can do with what you have now and the comment I made in post #3.
Ah, I can see where this is going. With three capacitors I can use my previous result to reduce it to an equivalent of two capacitors, and then repeat the argument.

Thanks for the help gneill!
 
  • #14
gneill
Mentor
20,913
2,862
For that, wouldn't it be necessary that C1+C2>1 ?
Why? For any positive values of ##C_1## and ##C_2##, can ##C_2## ever be greater than ##C_2 + C_1##?
 
  • #15
35,259
11,510
Why? For any positive values of ##C_1## and ##C_2##, can ##C_2## ever be greater than ##C_2 + C_1##?
@siddharth23: In addition to that, how would you compare capacitances to real numbers?
 
  • #16
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
I can show that it is the case for specific examples, but not a general case.
You have a capacitor, C, connected to a voltage source, V. Were you to add a second capacitor in series with C, some voltage will appear across that second capacitor, leaving less than V volts across C. Less voltage on C means there is less charge on C. Less common charge means the circuit's capacitance has been lowered, so must now be less than C.
 

Related Threads on Series Combo Of Capacitors, Smaller Than Single Capacitor

  • Last Post
Replies
3
Views
2K
Replies
1
Views
721
Replies
7
Views
1K
Replies
1
Views
2K
Replies
15
Views
189K
Replies
31
Views
1K
Replies
1
Views
5K
Replies
4
Views
3K
  • Last Post
Replies
4
Views
11K
Top