A 3.90-F capacitor is charged by a 12.0-V battery. It is disconnected from the b

Question

A 3.90-F capacitor is charged by a 12.0-V battery. It is disconnected from the battery and then connected to an uncharged 5.00-F capacitor

A). Determine the total stored energy before the two capacitors are connected. Express your answer using three significant figures and include the appropriate units.

B). Determine the total stored energy after they are connected. Express your answer using three significant figures and include the appropriate units.

C). What is the change in energy? Express your answer using three significant figures and include the appropriate units.

Explanation / Answer

here ,

A) initially total energy

U = 0.50 * C * V^2

U = 0.50 * 3.9 *10^-6 * 12^2

U= 2.81 *10-4 J

B)

let the charge on 5 uC is q

as the potential across 5 uC and 3.9 uC will be same when connected

(3.9 * 12 - q)/3.9 = q/5

solving for q

q = 26.3 uC

for the potential across capacitor

V = 26.3/5 = 5.26 V

total energy stored = 0.50 * (3.9 + 5) * 10^-6 * 5.26^2

total energy stored = 1.23 * 10-4 J

C)
change in energy = final energy - initial energy

change in energy = 1.23 * 10^-4 - 2.81 *10^-4

change in energy = -1.58 *10^-4 J

the change in energy is -1.58 *10-4 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.