Respuesta :
Answer:
Therefore,
The charge on the 4.20μF capacitor is 75.6μC.
The voltage across each capacitor in the parallel combination is 10.8 Volt.
Explanation:
Given:
Three capacitors having capacitance,
[tex]C_{1}=8.4\mu F\\C_{2}=8.40\mu F\\C_{3}=4.20\mu F\\[/tex]
are connected in series across a 36.0-V potential difference,
V= 36 V
To Find:
Qs = ? charge on the 4.20−μF
V = ? voltage across each capacitor in the parallel combination
Solution:
For Capacitor Series Combination we have,
[tex]\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}[/tex]
Substituting the values we get
[tex]\dfrac{1}{C_{s}}=\dfrac{1}{8.4}+\dfrac{1}{8.4}+\dfrac{1}{4.2}\\\\C_{s}=2.1\mu F[/tex]
As Capacitors are connected in series they have the same charge on each plates, and is given by
[tex]Q= C_{s}\times V[/tex]
Substituting the values we get
[tex]Q= 2.1\mu F\times 36=75.6\mu C\\Q=75.6\mu C[/tex]
Now when the plates are connected in Parallel we will have,
[tex]C_{p}=C_{1}+C_{2}+C_{3}[/tex]
Substituting the values we get
[tex]C_{p}=8.4+8.4+4.2=21\mu F[/tex]
As Capacitors are connected in Parallel they have the same Voltage on each plates, and is given by
[tex]V= \dfrac{Q_{1}+Q_{2}+Q_{3}}{C_{p}}[/tex]
Substituting the values we get
[tex]V=\dfrac{75.6+75.6+75.6}{21}=10.8\ V[/tex]
Therefore,
The charge on the 4.20μF capacitor is 75.6μC.
The voltage across each capacitor in the parallel combination is 10.8 Volt.
