Respuesta :
Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
The answers are
(a) the charge on capacitor 4.2μF is 71.4 μ C
(b) and the voltage across each capacitor in parallel combination is 10.2V
Given information:
Three capacitors C ₁= 8.4 μF, C₂ = 8.4 μF , C₃ = 4.2 μF connected in series with the potential difference V = 36 V
(a) The equivalent capacitance in series combination is given by
[tex]1/C = 1/C _1 + 1/C_2+ 1/C_3\\\\1/ C= 1/8.4 +1 / 8.4 + 1/4.2\\\\C=2.1 \mu F[/tex]
In series combination the charge on each capacitor is the same which can be calculated by the use of equivalent capacitance C as follows:
[tex]Q= CV\\\\Q= 2.1 \times 34 \mu C\\\\Q=71.4 \mu C[/tex]
so the charge on the capacitor 4.2μF is 71.4μC
(b) Now the capacitors are disconnected and they retain the charge but now they are connected in parallel combination.
the equivalent capacitance in parallel combination:
[tex]C'= C _1+C_2+C_3 \\ \\ C'= (8.4 + 8.4 + 4.2) \mu F\\\\C=21 \mu F[/tex]
In parallel combination, the voltage across the capacitors remains the same.
Now the total charge
Q'=3 Q
since the charge remains conserved
[tex]3Q=C'V'\\\\3\times71.4=21\times V'\\\\V'=10.2 V[/tex]is the voltage across each capacitor.
Learn more about equivalent capacitance:
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