Question

A 15-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.3 µF, 12.6 µF, and 31.2 µF. Find the voltage across the 31.2-µF capacitor.

_____________ V

Answer 1

The potential across the capacitor is 1.39 V.

Step-by-step explanation:

Given that,

Voltage = 15 V

First capacitance = 4.3μF

Second capacitance = 12.6μF

Third capacitance = 31.2 μF

We need to calculate the equivalent capacitance

Using formula of capacitance for series

`(1)/(C)=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`

Put the value into the formula

`(1)/(C)=(1)/(4.3)+(1)/(12.6)+(1)/(31.2)`

`(1)/(C)=(48455)/(140868)`

`C=2.907\ \mu F`

We need to calculate the charge

Using formula of charge

`q=CV`

Put the value into the formula

`q=2.907*15`

`q=43.605\ \mu C`

We need to calculate the potential difference

Using formula of potential difference

`\Delta V=(q)/(C)`

Put the value into the formula

`\Delta V=(43.605)/(31.2)`

`\Delta V=1.39\ V`

Hence, The potential across the capacitor is 1.39 V.