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

Answer:

The potential across the capacitor is 1.39 V.

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

$\dfrac{1}{C}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{4.3}+\dfrac{1}{12.6}+\dfrac{1}{31.2}$

$\dfrac{1}{C}=\dfrac{48455}{140868}$

$C=2.907\ \mu F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=2.907\times15$

$q=43.605\ \mu C$

We need to calculate the potential difference

Using formula of potential difference

$\Delta V=\dfrac{q}{C}$

Put the value into the formula

$\Delta V=\dfrac{43.605}{31.2}$

$\Delta V=1.39\ V$

Hence, The potential across the capacitor is 1.39 V.