Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 25, Problem 11P

Four capacitors are connected as shown in Figure P25.11. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking ΔVab = 15.0 V.

Figure P25.11

Chapter 25, Problem 11P, Four capacitors are connected as shown in Figure P25.11. (a) Find the equivalent capacitance between

(a)

Expert Solution
Check Mark
To determine
The equivalent capacitance between a and b .

Answer to Problem 11P

The equivalent capacitance between a and b is 5.96μF .

Explanation of Solution

Given information:

The four capacitors are connected in figure given below:

Physics for Scientists and Engineers, Chapter 25, Problem 11P , additional homework tip  1

Figure (1)

Explanation:

The capacitors 15.0μFand3.00μF are in series.

Formula to calculate the equivalent capacitance when they are connected in series.

CS=C15.0μF×C3.00μFC15.0μF+C3.00μF (1)

Here,

CS is the equivalent capacitance when they are connected in series.

Substitute 15.0μF for C15.0μF , 3.00μF for C3.00μF in equation (1) to find CS ,

CS=15.0μF×3.00μF15.0μF+3.00μF=2.50μF

The capacitors CS and 6.00μF are in parallel.

Formula to calculate the equivalent capacitance when they are connected in parallel.

CP=CS+C6.00μF (2)

Here,

CP is the equivalent capacitance when they are connected in parallel.

Substitute 2.50μF for CS , 6.00μF for C6.00μF in equation (1) to find CP ,

CP=2.50μF+6.00μF=8.50μF

The capacitors CP and 20.0μF are in series.

Formula to calculate the equivalent capacitance when they are connected in series.

Ceq=CP×C20.0μFCP+C20.0μF (3)

Here,

Ceq is the equivalent capacitance between a and b.

Substitute 8.50μF for CP , 20.0μF for C20.0μF in equation (1) to find Ceq ,

Ceq=8.50μF×20.0μF8.50μF+20.0μF=5.96μF

Thus, the equivalent capacitance between a and b is 5.96μF .

Conclusion:

Therefore, the equivalent capacitance between a and b is 5.96μF .

(b)

Expert Solution
Check Mark
To determine
The charge on each capacitor.

Answer to Problem 11P

The charge across capacitors 15.0μFand3.00μF is C2 is 26.3μC , the charge through 20.0μF is 89.5μC , the charge across 6.00μF capacitor is 63.2μC .

Explanation of Solution

Given information:

The voltage across a and b is 15.0V , the four capacitors are connected in figure given below:

Physics for Scientists and Engineers, Chapter 25, Problem 11P , additional homework tip  2

Explanation:

Formula to calculate the total charge in the circuit.

Qtotal=Ceq×ΔVab (4)

Here,

Qtotal is the total charge in the circuit.

ΔVab is the voltage across a and b.

Substitute 15.0V for ΔVab , 5.965μF for Ceq , in equation (3) to find VCP ,

Qtotal=5.965μF×15.0V=89.475μCQ20.0μF89.5μC

Thus, the total charge in the circuit and the charge through 20.0μF is 89.5μC .

Formula to calculate the potential drop across 20.0μF capacitor.

ΔV20.0μF=Q20.0μFC20.0μF (5)

Substitute 20.0μF for C20.0μF , 89.5μC for Q20.0μF , in equation (5) to find ΔV20.0μF ,

ΔV20.0μF=89.5μC20.0μF=4.475V

Thus, the potential drop across 20.0μF capacitor is 4.475V .

Formula to calculate the potential drop across 6.00μF capacitor.

ΔV6.00μF=ΔVabΔV20.0μF (6)

Substitute 15.0V for ΔVab , 4.475V for ΔV20.0μF , in equation (6) to find ΔV6.00μF ,

ΔV6.00μF=15.0V4.475V=10.525V

Thus, the potential drop across 6.00μF capacitor is 10.525V .

Formula to calculate the charge across 6.00μF capacitor.

Q6.00μF=C6.00μF×ΔV6.00μF (7)

Substitute 6.00μF for C6.00μF , 10.525V for ΔV6.00μF in equation (7) to find Q6.00μF ,

Q6.00μF=6.00μF×10.525V=63.15μC63.2μC

Thus, the charge across 6.00μF capacitor is 63.2μC .

The charge across capacitors 15.0μFand3.00μF will be same.

Calculate the charge for the capacitor 15.0μFand3.00μF .

Q15.0μF=Q3.00μF=Q20.0μFQ6.00μF (8)

Substitute 89.5μC for Q20.0μF , 63.2μC for Q6.00μF in equation (8) to find Q15.0μFandQ3.00μF ,

Q15.0μFand3.00μF=89.5μC63.2μC=26.3μC

Thus, the charge across capacitors 15.0μFand3.00μF is C2 is 26.3μC .

Conclusion:

Therefore, the charge across capacitors 15.0μFand3.00μF is C2 is 26.3μC , the charge through 20.0μF is 89.5μC , the charge across 6.00μF capacitor is 63.2μC .

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Chapter 25 Solutions

Physics for Scientists and Engineers

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