Part A A 4.40 µF capacitor is charged by a 12.0 V battery. It is disconnected from the battery and then connected to an uncharged 5.10 µF capacitor (see the figure (Figure 1)). Determine the total stored energy before the two capacitors are connected. U1 = 3.17×10-4 J Submit Previous Answers Correct Part B Determine the total stored energy after they are connected. nνα ΑΣφ ? U2 = J Submit Request Answer Figure < 1 of 1 > Part C C What is the change in energy? ηνα ΑΣφ
Part A A 4.40 µF capacitor is charged by a 12.0 V battery. It is disconnected from the battery and then connected to an uncharged 5.10 µF capacitor (see the figure (Figure 1)). Determine the total stored energy before the two capacitors are connected. U1 = 3.17×10-4 J Submit Previous Answers Correct Part B Determine the total stored energy after they are connected. nνα ΑΣφ ? U2 = J Submit Request Answer Figure < 1 of 1 > Part C C What is the change in energy? ηνα ΑΣφ
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Transcribed Image Text:Part A
A 4.40 µF capacitor is charged by a 12.0 V battery. It is disconnected
from the battery and then connected to an uncharged 5.10 µF
capacitor (see the figure (Figure 1)).
Determine the total stored energy before the two capacitors are connected.
U1 = 3.17x10-4 J
Submit
Previous Answers
Correct
Part B
Determine the total stored energy after they are connected.
να ΑΣφ
?
U2 =
J
Submit
Request Answer
Figure
< 1 of 1 >
Part C
What is the change in energy?
V ΑΣφ
?
U2 – U =
J
V
C2
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