Question

A capacitor of capacitance 5⋅00 µF is charged to 24⋅0 V and another capacitor of capacitance 6⋅0 µF is charged to 12⋅0 V. (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go?

Answer 1

a) Energy stored in each capacitor = 0.432 mJ

b) Charges on the capacitors = 21.8 & 26.2 uC

c) Loss of electrostatic energy = 1.77 mJ

d) The energy dissipated as heat.

Explanation:

Given C1= 5⋅00 µF & C2 =6⋅00 µF  

V1 =24 Volts & V2 = 12 Volts

Q1 = C1V1 = $5 \times 24$ = 120 µC

Q2 = C2V2 = $6 \times 12$ = 72 µC

Energy stored in the first capacitor:

 U = ½ CV2

= 1440 J = 1.44 mJ

U2 = ½ CV2

= 432 J = 0.432 mJ

b) The capacitors are connected to each other in such a way that the positive plate of the first capacitor is connected to the negative plate of the second capacitor and vice versa.

Net charge Q= 120 -72 = 48

Consider (V) be the common potential of the two capacitors

V= Q/C1+C2 = 48/5+6 = 4.36V

New Charges Q1 = C1V = $5 \times 4.36$ = 21.8 µC

Q2 = C2V = $6 \times 4.36$ = 26.2 µC

C) Given U = ½ CV2

= ½ (C1+C2)

= 0.5 (4.36)2 + (5+6)

$104.5 \times 10-6 J$ = 0.1045 mJ

Loss of energy = 1.873 – 0.1045 = 1.7678= 1.77 mJ

d) The energy is dissipated as heat.

Answer 2

Explanation:

a) Energy stored in each capacitor = 0.432 mJ

b) Charges on the capacitors = 21.8 & 26.2 uC

c) Loss of electrostatic energy = 1.77 mJ

d) The energy dissipated as heat.

Explanation:

Given C1= 5⋅00 µF & C2 =6⋅00 µF

V1 =24 Volts & V2 = 12 Volts

Q1 = C1V1 = 5 \times 245×24 = 120 µC

Q2 = C2V2 = 6 \times 126×12 = 72 µC

Energy stored in the first capacitor:

U = ½ CV2

= 1440 J = 1.44 mJ

U2 = ½ CV2

= 432 J = 0.432 mJ

b) The capacitors are connected to each other in such a way that the positive plate of the first capacitor is connected to the negative plate of the second capacitor and vice versa.

Net charge Q= 120 -72 = 48

Consider (V) be the common potential of the two capacitors

V= Q/C1+C2 = 48/5+6 = 4.36V

New Charges Q1 = C1V = 5 \times 4.365×4.36 = 21.8 µC

Q2 = C2V = 6 \times 4.366×4.36 = 26.2 µC

C) Given U = ½ CV2

= ½ (C1+C2)

= 0.5 (4.36)2 + (5+6)

104.5 \times 10-6 J104.5×10−6J = 0.1045 mJ

Loss of energy = 1.873 – 0.1045 = 1.7678= 1.77 mJ

d) The energy is dissipated as heat.