Question

three capacitors of different values are given. how should the be combined to give min and max capacitance?

Answer 1

Answer:

The parallel connection gives us the maximum capacitance

Explanation:

Connection of Capacitors

Two capacitors can be connected in series or in parallel. If the capacitances are C1 and C2, and they are connected in series, then the equivalent capacitance is given by

$\displaystyle \frac{1}{C_e}=\frac{1}{C_1}+\frac{1}{C_2}$

Computing Ce

$\displaystyle C_e=\frac{C_1C_2}{C_1+C_2}$

If the are connected in parallel, the equivalent capacitance is

$C_e=C_1+C_2$

Now it's easy to prove that  

$\displaystyle C_1+C_2>\frac{C_1C_2}{C_1+C_2}$

Operating

$(C_1+C_2)(C_1+C_2)>C_1C_2$

Note both sides of the inequality are the product of two numbers. The right side multiplies C1+C2 by itself, and C1+C2 is greater than C1 or C2 alone, thus the product it greater too.  This analysis can be extended to more than two capacitors.

Thus, the parallel connection gives us the maximum capacitance