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

[tex]\dfrac{1}{C}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}[/tex]

Put the value into the formula

[tex]\dfrac{1}{C}=\dfrac{1}{2.0}+\dfrac{1}{3.0}+\dfrac{1}{6.0}[/tex]

[tex]\dfrac{1}{C}=1[/tex]

[tex]C=1\ \mu F[/tex]

We need to calculate the charge

Using formula of charge

[tex]q=CV[/tex]

Put the value into the formula

[tex]q=1\times12.0[/tex]

[tex]q=12\ \mu C[/tex]

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

[tex]V_{1}=\dfrac{q}{C_{1}}[/tex]

Put the value into the formula

[tex]V_{1}=\dfrac{12}{2.0}[/tex]

[tex]V_{1}=6.0 V[/tex]

We need to calculate the voltage across second capacitor

Using formula of voltage

[tex]V_{2}=\dfrac{q}{C_{2}}[/tex]

Put the value into the formula

[tex]V_{2}=\dfrac{12}{3.0}[/tex]

[tex]V_{2}=4.0 V[/tex]

We need to calculate the voltage across third capacitor

Using formula of voltage

[tex]V_{3}=\dfrac{q}{C_{3}}[/tex]

Put the value into the formula

[tex]V_{3}=\dfrac{12}{6.0}[/tex]

[tex]V_{3}=2.0 V[/tex]

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

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