For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor. Figure P16.41 Problems 41 and 60.

Question

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Answer 1

Textbook Question

Chapter 16, Problem 41P

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Chapter 16, Problem 41P, For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the

Figure P16.41 Problems 41 and 60.

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

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Students have asked these similar questions

Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?

* Coulomb's Law Example Three charges are positioned as seen below. Charge 1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is - 6.0MC. What is the magnitude and the direction of the force on charge 2 due to charges 1 and 3? 93 kq92 F == 2 r13 = 0.090m 91 r12 = 0.12m 92 Coulomb's Constant: k = 8.99x10+9 Nm²/C² ✓

Make sure to draw a Free Body Diagram as well

Answer 2

Textbook Question

Chapter 16, Problem 41P

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Chapter 16, Problem 41P, For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the

Figure P16.41 Problems 41 and 60.

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

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Answer 3

Textbook Question

Chapter 16, Problem 41P

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Chapter 16, Problem 41P, For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the

Figure P16.41 Problems 41 and 60.

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

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Answer 4

Textbook Question

Chapter 16, Problem 41P

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Chapter 16, Problem 41P, For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the

Figure P16.41 Problems 41 and 60.

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Answer 8

(a)

Expert Solution

To determine

The equivalent capacitance.

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The equivalent capacitance.

Answer 13

Answer to Problem 41P

The equivalent capacitance is $3.33 μF$ .

Answer 14

Explanation of Solution

The capacitors $6.00 μF$ and $3.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpt=C6.00 μFC3.00 μFC6.00 μF+C3.00 μF$

The capacitors $2.00 μF$ and $4.00 μF$ are connected in series combination. The equivalent capacitance is,

$Cpb=C2.00 μFC4.00 μFC2.00 μF+C4.00 μF$

The capacitances $Cpt$ and $Cpb$ are in parallel combination. The total equivalent capacitance is,

$Ceq=Cpt+Cpb$

Therefore,

$Ceq=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)+(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ , $2.00 μF$ for $C2.00 μF$ and $4.00 μF$ for $C4.00 μF$

$Ceq=[(6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF)]+[(2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF)]=3.33 μF$

On Re-arranging,

$Ceq=8.00 μF3=2.67 μF$

Conclusion:

The equivalent capacitance is $3.33 μF$ .

Answer 15

(b)

Expert Solution

To determine

The charge on each capacitor.

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The charge on each capacitor.

Answer 20

Answer to Problem 41P

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Answer 21

Explanation of Solution

Formula to calculate the charge on $6.00 μF$ and $3.00 μF$ capacitors is,

$Q1=CptV$

Therefore,

$Q1=(C6.00 μFC3.00 μFC6.00 μF+C3.00 μF)V$

Substitute $6.00 μF$ for $C6.00 μF$ , $3.00 μF$ for $C3.00 μF$ and 90.0 V for V.

$Q1=((6.00 μF)(3.00 μF)(6.00 μF)+(3.00 μF))(90.0 V)=180 μC$

Formula to calculate the charge on $2.00 μF$ and $4.00 μF$ capacitors is,

$Q2=CpbV$

Therefore,

$Q2=(C2.00 μFC4.00 μFC2.00 μF+C4.00 μF)V$

Substitute $2.00 μF$ for $C2.00 μF$ , $4.00 μF$ for $C4.00 μF$ $C3.00 μF$ and 90.0 V for V.

$Q2=((2.00 μF)(4.00 μF)(2.00 μF)+(4.00 μF))(90.0 V)=120 μC$

Conclusion:

The charge on $6.00 μF$ and $3.00 μF$ capacitors is $180 μC$

The charge on $2.00 μF$ and $4.00 μF$ capacitors is $120 μC$

Answer 22

(c)

Expert Solution

To determine

The potential difference on each capacitor.

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The potential difference on each capacitor.

Answer 27

Explanation of Solution

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Formula to calculate the potential difference on $2.00 μF$ and $3.00 μF$ capacitors is,

$V=Q2C2.00 μF$

Substitute $120 μC$ for $Q2$ and $2.00 μF$ for $C2.00 μF$

$V=120 μC2.00 μF=60 V$

Formula to calculate the potential difference on $6.00 μF$ and $4.00 μF$ capacitors is,

$V=Q2C6.00 μF$

Substitute $180 μC$ for $Q1$ and $6.00 μF$ for $C6.00 μF$

$V=180 μC6.00 μF=30 V$

Conclusion:

The potential difference on $2.00 μF$ and $3.00 μF$ capacitors is 60 V

The potential difference on $6.00 μF$ and $4.00 μF$ capacitors is 30 V

Answer 28

Ch. 16.1 - If an electron is released from rest in a uniform...Ch. 16.1 - If a negatively charged particle is placed at rest...Ch. 16.1 - Figure 16.3 is a graph of an electric potential as...Ch. 16.1 - If a negatively charged particle is placed at...Ch. 16.2 - Consider a collection of charges in a given region...Ch. 16.2 - A spherical balloon contains a positively charged...Ch. 16.3 - An electron initially at rest accelerates through...Ch. 16.6 - A capacitor is designed so that one plate is large...Ch. 16.7 - A parallel-plate capacitor is disconnected from a...Ch. 16.8 - A fully charged parallel-plate capacitor remains...

For the system of capacitors shown in Figure P16.41, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor. Figure P16.41 Problems 41 and 60.

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Answer to Problem 41P

Explanation of Solution

Answer to Problem 41P

Explanation of Solution

Explanation of Solution

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