Two uncharged capacitors, C1 (1.6?F) and C2 (7.4?F), are connected in parallel t
ID: 2033549 • Letter: T
Question
Two uncharged capacitors, C1 (1.6?F) and C2 (7.4?F), are connected in parallel to a battery with an EMF of 14 V. After a very long time, they are disconnected and carefully (so as not to disturb the charge on each plate) reconnected to each other, positive plate to negative plate, and negative plate to positive plate. There is no battery in this new circuit. Below is a diagram of the capacitors immediately after they are reconnected, BEFORE any charges are transfered between plates. C2 A)How much excess charge exists in each half of the circuit? Submit Answer Tries 0/99 B) After the charges have balanced out according to the rules of energy and charge conservation, which diagram best represents the plate configuration? Ct C2 C2 iv) ou are correct. Your receipt no. is 155-5751?) Previous TriesExplanation / Answer
C1 = capacitance of capacitor 1 = 1.6 uF
C2 = Capacitance of capacitor 2 = 7.4 uF
V = battery Voltage = 14 Volts
Q1 = charge stored in capacitor 1 = C1 V = 1.6 x 14 = 22.4 uC
Q2 = charge stored in capacitor 2 = C2 V = 7.4 x 14 = 103.6 uC
A)
excess charge is given as
Q = Q2 - Q1 = 103.6 - 22.4 = 81.2 uC
D)
Q1' + Q2' = Q2 - Q1 = 103.6 + 22.4 = 126 uC
Q1'/C1 = Q2'/C2
(126 - Q2' )/1.6 = Q2'/7.4
Q2' = 103.6 uC
Q1' = 126 - 103.6 = 22.4 uC
potential difference across each capacitor= V' = Q2'/7.4 = 103.6/7.4 = 14 Volts
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.