Question

A 330 pF capacitor and a 220 pF capacitor in series are each connected across a 6 V dc source. What is the voltage across the 330 pF capacitor?

Answer 1

`2.4"V"`

Explanation:

For capacitors in series like this the total charge stored by each capacitor is the same.

The capacitance `C`, charge stored `Q` and the voltage `V` are related by:

`C=Q/V`

So:

`Q=CV`

Let `C_1=330"pF"`

and `C_2=220"pF"`

So:

`C_1V_1=C_2V_2` `" "color(red)((1))`

The voltage drop across the 2 capacitors = 6V so we can write:

`V_1+V_2=6` `" "color(red)((2))`

We have 2 simultaneous equations here so from `color(red)((2))` we can write:

`V_2=(6-V_1)`

We can now substitute this expression for `V_2` into `color(red)((1))rArr`

`C_1V_1=C_2(6-V_1)`

`C_1V_1=6C_2-C_2V_1`

`C_1V_1+C_2V_1=6C_2`

`V_1(C_1+C_2)=6C_2`

`V_1=(6C_2)/((C_1+C_2))`

`V_1=(6xx220xxcancel(10^(-12)))/((330+220)xxcancel(10^(-12)))`

`V_1=1320/550=2.4"V"`

• which is the voltage drop across the `330"pF"` capacitor.

This also means that `V_2=6-2.4=3.6"V"`