Respuesta :
Answer:
Explanation:
combined capacitor of parallel capacitors
1.51 + 2.99
= 4.5 μF
Resultant capacitance of all capacitor
= 4.5 x 4.41 / ( 4.5 +4.41) μF
= 2.22 μF
Charge on series capacitor
= C X V
= 16.5 X 2.22 μC
= 36.63 μC
Potential on series capacitor (4.41 μF)
= Q / C
= 36.63 / 4.41
= 8.3 V
Potential on each of parallel capacitor
= 16.5 - 8.3
= 8.2 V
Charge on 2.99 μF capacitor
= 2.99 x 8.2 = 24.52 C
Charge on 1.51 μF capacitor
= 1.51 x 8.2 =12.38 C
=
Answer:
Explanation:
C1 = 1.51 μF
C2 = 2.99 μF
C3 = 4.41 μF
V = 16.5 V
Here C1 and C2 are connected in parallel.
Cp = C1 + C2 = 1.51 + 2.99 = 4.5 μF
C3 and Cp are in series
The equivalent capacitance is
[tex]C=\frac{C_{p}C_{3}}{C_{p}+C_{3}}[/tex]
[tex]C=\frac{4.5\times 4.41}{4.5+4.41}[/tex]
C = 2.23 μF
Let the total charge is q.
q = C x V = 2.23 x 16.5 = 36.8 μC
Voltage on C3, V3 = q / C3 = 36.8 / 4.41 = 8.35 V
Voltage on C1, V1 = Voltage on C2, V2 = V - V3 = 16.5 - 8.35 = 8.15 V
Charge on C1, q1 = C1 x V1 = 1.51 x 8.15 = 12.3 μC
Charge on C2, q2 = C2 x V2 = 2.99 x 8.15 = 24.4 μC
Charge on C3, q3 = q = 36.8 μC
