Respuesta :
Answer:
Hence,
The voltage across the 10.00 µF capacitor is 18.75 Volts.
Explanation:
Given:
A 3.00 µF and a 5.00 µF capacitor are connected in series across a 30.0 V battery,
[tex]C_{1}=3\mu F\\C_{2}=5\mu F\\[/tex]
V = 30.0 V
To Find:
V₁ = ? (voltage across the 10.00 µF capacitor)
Solution:
For Capacitor Series Combination we have,
[tex]\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}[/tex]
Substituting the values we get
[tex]\dfrac{1}{C_{s}}=\dfrac{1}{3}+\dfrac{1}{5}\\C_{s}=1.875\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= 1.875\mu F\times 30=56.25\mu C\\Q=56.25\mu C[/tex]
Also in Series voltage will be different across C₁ and C₂,
Therefore voltage across C₁ will be
[tex]V_{1}= \dfrac{Q}{C_{1}}[/tex]
Substituting the values we get
[tex]V_{1}=\dfrac{56.25}{3}=18.75\ Volt[/tex]
Now according to the given condition,
10.00 µF capacitor is then connected in parallel across the 3.00 µF capacitor,
Therefore we know in Parallel, Voltage remain Same, Hence Voltage across 10.00 µF capacitor will be same as that of voltage across V₁.
Hence,
The voltage across the 10.00 µF capacitor is 18.75 Volts.
