Respuesta :
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.
