Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 18, Problem 129P

(a)

To determine

Find the total energy stored in two capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 129P

The total energy stored in two capacitor is 175 μJ.

Explanation of Solution

Write the equation for energy.

U=12CV2 (I)

Here, U is the energy, C is the capacitance and V is potential difference.

Conclusion:

For 2.00 μF capacitor, substitute 5.00 V for V and 2.00 μF for C in equation I.

U1=12(2.00 μF)(5.00 V)2=25 μJ

For 3.00 μF capacitor, substitute 10.0 V for V and 3.00 μF for C in equation I.

U2=12(3.00 μF)(10.0 V)2=150 μJ

Then the total energy stored is U=U1+U2=25 μJ+150 μJ=175 μJ

Thus, the total energy stored in two capacitor is 175 μJ.

(b)

To determine

Find the total energy in the two capacitor and charge on each capacitor after disconnected from batteries.

(b)

Expert Solution
Check Mark

Answer to Problem 129P

The total energy in the two capacitor is 160 μJ and charge on each capacitor after disconnected from batteries are 16.0 μC and 24.0 μC.

Explanation of Solution

Write the equation for total charge.

Q=C2V2i+C3V3i (II)

Here, Q is the total charge, C2 is the capacitance of 2.00 μF capacitor, C3 is the capacitance of 3.00 μF capacitor, V2i is voltage across 2.00 μF capacitor and V3i is voltage across 3.00 μF capacitor.

Since the two capacitors are connected together the voltage across them is same, then the charge on each capacitor are,

Q2f=C2V and Q3f=C3V

From the above two equation eliminate V.

Q2fC2=Q3fC3 (III)

The total charge is written as,

Q=Q2f+Q3f (IV)

Substitute the value of Q3f from the above equation to equation III.

Q2fC2=Q3fC3=QQ2fC3

Solve the above equation to get Q2f.

Q2f(1C2+1C3)=QC3Q2f=QC3(1C2+1C3)

Q2f=QC3C2+1 (V)

Write the equation for total charge.

Utotal=12C2V2+12C3V2 (VI)

Conclusion:

Substitute 2.00 μF for C2, 3.00 μF for C3, 5.00 V for V2i and 10.0 V for V3i  in equation II.

Q=(2.00 μF)(5.00 V)+(3.00 μF)(10.0 V)=40.0 μC

Substitute 40.0 μC for Q, 2.00 μF for C2 and 3.00 μF for C3 in equation V.

Q2f=40.0 μC3.00 μF2.00 μF+1=16.0 μC

Substitute 16.0 μC for Q2f and 40.0 μC for Q in equation IV.

Q3f=40.0 μC16.0 μC=24.0 μC

Find the value of V.

V=Q2fC2=16.0 μC2.00 μF=8.00 V

Substitute 8.00 V for V, 2.00 μF for C2 and 3.00 μF for C3 in equation VI

Utotal=12(2.00 μF)(8.00 V)2+12(3.00 μF)(8.00 V)2=160 μJ

Therefore, the total energy in the two capacitor is 160 μJ and charge on each capacitor after disconnected from batteries are 16.0 μC and 24.0 μC.

(c)

To determine

Explain the reason for missing energy.

(c)

Expert Solution
Check Mark

Answer to Problem 129P

The missing energy is due to the appearance of internal energy in the wires.

Explanation of Solution

The missing energy is due to the internal energy caused by the wires connecting the capacitors during the current flowing time.

Conclusion:

Therefore, the missing energy is due to the appearance of internal energy in the wires.

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