The figure below shows three capacitors with capacitances C A = 1.00 µF, C B = 1
ID: 1423952 • Letter: T
Question
The figure below shows three capacitors with capacitances CA = 1.00 µF, CB = 1.80 µF, and CC = 4.60 µF connected to a3.00-V battery.
The figure below shows three capacitors with capacitances CA : 1.00 F, 3.00-V battery. 1.80 and CC : 4.60-F connected to a CA (a) What is the equivalent capacitance of the three capacitors? 5 Your response differs from the correct answer by more than 100%. F (b) What charge is stored in each of the capacitors? QA 5 1.86 Your response differs from the correct answer by more than 100%. c B 5 3.36 Your response differs from the correct answer by more than 10%. Double check your calculations. C 5 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. HCExplanation / Answer
part A
Ca and Cb arein parallel combination therefore net capacitance if these two capacitors is = Ca + Cb = 2.8 µF
now this combination is inseries with the Cc capacitor therefire net capcitance will be = 2.8 µF || 4.6 µF = 1.74 µF ANSWER
using this net capacitance we can find net charge stored in all the capacitors which is equal to Q = CV
Qnet = 3 V * 1.74 µF = 5.22 µC
let Qa = Ca*V1 and Qb = Cb*V1 since Ca and Cb are in parallel therefore potential difference across both of them will be same and Qc = Cc*V2
Let's first find the potential difference across these capacitors
for finding the potential difference across each capacitors
firstly combine Ca and Cb which is equal to 2.8 µF now the circuit is as follows:
a battery in series with the 4.6 µF capacitor and a 2.8 µF capacitor there
using the equations written above we can say that '
V1 + V2 = 3 Volts..........................(1)
also Qa + Qb + Qc = Qnet
which is equal to (Ca + Cb)*V1 + Cc*V2 = Cnet*V
2.8 µF * V1 + 4.6 µF*V2 = 5.22 µC..............................(2)
on solving equation 1 and 2 we get
V1 = 4.766 V V2 = -1.766 V
charges on Ca Qa = Ca * V1 = 4.766 µC
charges on Cb Qb = Cb*V1 = 8.57988 µC
charge on Cc Qc = Cc*V2 = -8.12636 µC