# Series Combo Of Capacitors, Smaller Than Single Capacitor

1. Jan 27, 2015

### BOAS

1. The problem statement, all variables and given/known data

"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.

2. Relevant equations

3. The attempt at a solution

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.

Jacob.

2. Jan 27, 2015

### CWatters

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

3. Jan 27, 2015

### Staff: Mentor

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. Jan 27, 2015

### BOAS

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

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

5. Jan 27, 2015

### Staff: Mentor

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

6. Jan 27, 2015

### BOAS

Okay,

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

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

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

7. Jan 27, 2015

### Staff: Mentor

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

8. Jan 27, 2015

### BOAS

Not really...

9. Jan 27, 2015

### Staff: Mentor

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

10. Jan 27, 2015

### BOAS

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. Jan 27, 2015

### Staff: Mentor

See what you can do with what you have now and the comment I made in post #3.

12. Jan 27, 2015

### siddharth23

For that, wouldn't it be necessary that C1+C2>1 ?

13. Jan 27, 2015

### BOAS

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. Jan 27, 2015

### Staff: Mentor

Why? For any positive values of $C_1$ and $C_2$, can $C_2$ ever be greater than $C_2 + C_1$?

15. Jan 27, 2015

### Staff: Mentor

@siddharth23: In addition to that, how would you compare capacitances to real numbers?

16. Jan 27, 2015

### Staff: Mentor

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.