Question

You a building a circuit that requires a 150 pF capacitor, but you only have a drawer full of 100 pF capacitors.

(How) can you make a 150 pF equivalent capacitor?

Answer 1

To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.

When capacitors are connected in series, they consecutively add their capacitance.

However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is
`\frac {1} {C}` where, C is the capacitance.

For the given case, it is best to connect two of these capacitors in series and one in parallel, so

We have three 100 pF capacitors, then

`C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\`

Here we can see how two capacitors 1 and 2 are in series and the third in parallel.

`(1)/(C_(serie)) = (1)/(100pF)+(1)/(100pF)`

`(1)/(C_(serie)) = (1)/(50pF)`

Investing equality

`C_(serie) = 50pF`

Adding it in parallel, then

`C_(total) = C_(serie) +C_3`

`C_(total) = 50pF+100pF`

`C_(total) = 150pF`