Respuesta :
Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ = [tex]\frac{C_{3}C_{4} }{C_{3} + C_{4} }[/tex]
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ = [tex]\frac{4.25\times3.98 }{4.25 + 3.98 }[/tex]
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = [tex]\frac{Q}{C_{4} }[/tex] = [tex]\frac{35.46\times10^{-6} }{3.98\times10^{-6}}[/tex] = 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC
