The figure below shows three capacitors with capacitances C A = 1.00 µF, C B = 2
ID: 1429840 • Letter: T
Question
The figure below shows three capacitors with capacitances
CA = 1.00 µF,
CB = 2.40 µF,
and
CC = 3.40 µF
connected to a 3.00-V battery.
(a) What is the equivalent capacitance of the three capacitors?
µF
(b) What charge is stored in each of the capacitors?
(c) What is the potential difference across each of the capacitors?
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. µC QB= µC QC= µC
Explanation / Answer
(a) CA and CB are in series so there combination C' = 1 + 2.40 = 3.40 µF
Now C' is in series with Cc So net C = (3.40 x 3.40) / (3.40 + 3.40) = 1.7 µF
(c)C' and Cc are in series and are of equal magnitude So voltage is equally divided on both capacitors
V across Cc = 1.5 V
V across C' = 1.5 V (means V across CA and CB = V across C' because in parallel V kis same = 1.5 V)
(b) Q = CV
QA = 1 x 1.5 = 1.5 µC
QB = 2.40 x 1.5 = 3.6 µC
QC = 3.40 x 1.5 = 5.1 µC