1.) A circuit is built from three capacitors and a battery with values: C1=3150n
ID: 2067153 • Letter: 1
Question
1.) A circuit is built from three capacitors and a battery with values:C1=3150nanoFarad
C2=500nanoFarad
C3=3000nanoFarad
V=5V
Capacitors C1 and C2 are connected in series. This branch is connected in parallel to capacitor C1, and both these branches are connected in parallel to the battery V. (See Figure 27.26 pg 895 Lea & Burke.) What is the charge on each capacitor, and what is the total energy stored on all three capacitors?
Q1= C
Q2= C
Q3= C
U= J
2.) Capacitors C1=3400nanoFarad and C2=3350nanoFarad are initially connected in parallel and hold a combined charge of Q0=992nanoCoulombs. Find the charge on each capacitor and the energy stored in the system.
Q1= nanoCoulombs
Q2= nanoCoulombs
Uinitial= nanoJoules
The capacitors are then disconnected and reconnected with their terminals reveresed, so that the positively charged side of C1 is now connected to the negatively charged side of C2 and vice versa. Find the combined charge and energy stored in the final state.
Qfinal= nanoCoulombs
Ufinal= nanoJoules
Explanation / Answer
(1) so Q is just capacitance times potential. so Q_1 is 1.575x10^-5 C
Q_2 is 2.5x10^-6 C
Q_3 is 1.5x10^-5 C
U is 0.5CV^2 so I get 3.43x10^-6 for total capacitance and divide that by 2 and you get 1.716x10^-6. Now multiply that to V^2 which is 25 and you get 4.29x10^-5 Joules.
(2) I assume that we are using the same from the previous problem potential which is 5 volts.
Q_1 is 17000.0 nano Coulombs
Q_2 is 16750.0 nano Coulombs
U init. is 12400.0 nano Joules
(3) stuff doesnt change if the terminal is flipped. So Q final is 992 nano Coulombs and U final is 12400.0 nano Joules.
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