Question

three capacitors are connected in series to an ac voltage source with amplitude 12.0 V and frequency 6.3 kHz. What are the peak voltages across each capacitor

Answer 1

Answer:

The peak voltages across each capacitor are 6.0 V, 4.0 V and 2.0 V.

Explanation:

Given that,

Amplitude = 12.0 V

Frequency = 6.3 kHz

Suppose, Three capacitors are 2.0 μF, 3.0 μF and 6.0 μF

We need to calculate the equivalent capacitor

Using formula of series

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

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{2.0}+\dfrac{1}{3.0}+\dfrac{1}{6.0}$

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

$C=1\ \mu F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=1\times12.0$

$q=12\ \mu C$

The charge will be same in each capacitor because the capacitors are connected in series.

We need to calculate the voltage across first capacitor

Using formula of voltage

$V_{1}=\dfrac{q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{12}{2.0}$

$V_{1}=6.0 V$

We need to calculate the voltage across second capacitor

Using formula of voltage

$V_{2}=\dfrac{q}{C_{2}}$

Put the value into the formula

$V_{2}=\dfrac{12}{3.0}$

$V_{2}=4.0 V$

We need to calculate the voltage across third capacitor

Using formula of voltage

$V_{3}=\dfrac{q}{C_{3}}$

Put the value into the formula

$V_{3}=\dfrac{12}{6.0}$

$V_{3}=2.0 V$

Hence, The peak voltages across each capacitor are 6.0 V, 4.0 V and 2.0 V.