Series Combo Of Capacitors, Smaller Than Single Capacitor

[BOAS]

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.

 

[CWatters]

Perhaps start by showing that 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ etc

 

[gneill]

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).

 

[BOAS]

Perhaps start by showing that 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ etc

Ok,

$V = \frac{Q}{C}$

So the voltages on the individual capacitors is given by;

$V_{1} = \frac{Q}{C_{1}}$,

$V_{2} = \frac{Q}{C_{2}}$,

$V_{3} = \frac{Q}{C_{3}}$ etc...

The total voltage is $V_{tot} = V_{1} + V_{2} + V_{3} + ...$

$V = \frac{Q}{C_{eq}} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} + ...$

Dividing through by $Q$ yields

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + ...$

 

[gneill]

Okay, now using what you've done, write and simplify the equation for the equivalent capacitance of just two capacitors in series.

 

[BOAS]

Okay, now using what you've done, write and simplify the equation for the equivalent capacitance of just two capacitors in series.

Okay,

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$C_{1}C_{2} = C_{eq}C_{2} + C_{eq}C_{1}$

$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}$

 

[gneill]

Excellent. Now does that last expression suggest anything to you about showing that $C_{eq} < C_1$ and $C_{eq} < C_2$?

 

[BOAS]

Excellent. Now does that last expression suggest anything to you about showing that $C_{eq} < C_1$ and $C_{eq} < C_2$?

Not really...

 

[gneill]

What if I write the expression this way:
$C_{eq} = C_1 \frac{C_2}{C_2 + C_1}$

 

[BOAS]

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?.

 

[gneill]

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.

 

[siddharth23]

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 ?

 

[BOAS]

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!

 

[gneill]

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 $C_2 + C_1$?

 

[mfb]

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?

 

[NascentOxygen]

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.