College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 16, Problem 48P

Four capacitors are connected as shown in Figure P16.48.

(a) Find the equivalent capacitance between points a and b.

(b) Calculate the charge on each capacitor, taking ΔVab = 15.0 V.

Chapter 16, Problem 48P, Four capacitors are connected as shown in Figure P16.48. (a) Find the equivalent capacitance between

Figure P16.48

(a)

Expert Solution
Check Mark
To determine
The equivalent capacitance.

Answer to Problem 48P

The equivalent capacitance is 5.96μF .

Explanation of Solution

The capacitors 15.0μF and 3.00μF are connected in series combination. The equivalent capacitance is,

Cs=C15.0μFC3.00μFC15.0μF+C3.00μF

The capacitors Cs and 6.00μF are connected in parallel combination. The equivalent capacitance is,

Cp=Cs+(6.00μF)

The capacitance Cp and 20.0μF are connected in series combination. The total equivalent capacitance is,

Ceq=(C15.0μFC3.00μFC15.0μF+C3.00μF+C6.00μF)(C20.0μF)(C15.0μFC3.00μFC15.0μF+C3.00μF+C6.00μF)+(C20.0μF)

Substitute 15.0μF for C15.0μF , 3.00μF for C3.00μF , 6.00μF for C6.00μF and 20.0μF for C20.0μF

Ceq=((15.0μF)(3.00μF)(15.0μF)+(3.00μF)+6.00μF)(20.0μF)((15.0μF)(3.00μF)(15.0μF)+(3.00μF)+6.00μF)+(20.0μF)=5.96μF

Conclusion:

The equivalent capacitance is 5.96μF .

(b)

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

Answer to Problem 48P

The charge on 20.0μF capacitor is 89.4μC

The charge on 6.00μF capacitor is 63.0μC

The charge on 15.0μF and 3.00μF capacitors is 26.3μC

Explanation of Solution

Formula to calculate the charge on 20.0μF capacitor is,

Q20=Ceq(ΔV)

Substitute 5.96μF for Ceq and 15.0 V for V.

Q20=(5.96μF)(15.0V)=89.4μC

Formula to calculate the charge on 20.0μF capacitor is,

Q6=Q20(C6.00μFCs+C6.00μF)

Therefore,

Q6=Q20(C6.00μFC15.0μFC3.00μFC15.0μF+C3.00μF+C6.00μF)

Substitute15.0 V for V, 3.00μF for C3.00μF , 6.00μF for C6.00μF and 20.0μF for C20.0μF

Q6=(89.4μC)(6.00μF(15.0μF)(3.00μF)(15.0μF)+(3.00μF)+6.00μF)=63.0μC

Explanation:

Formula to calculate the charge on charge on 15.0μF and 3.00μF capacitors is,

Q=Q20(CsCs+C6.00μF)

Therefore,

Q6=Q20(C15.0μFC3.00μFC15.0μF+C3.00μFC15.0μFC3.00μFC15.0μF+C3.00μF+C6.00μF)

Substitute15.0 V for V, 3.00μF for C3.00μF , 6.00μF for C6.00μF and 20.0μF for C20.0μF

Q6=(89.4μC)((15.0μF)(3.00μF)(15.0μF)+(3.00μF)(15.0μF)(3.00μF)(15.0μF)+(3.00μF)+6.00μF)=26.3μC

Conclusion:

The charge on 20.0μF capacitor is 89.4μC .

The charge on 6.00μF capacitor is 63.0μC .

The charge on 15.0μF and 3.00μF capacitors is 26.3μC .

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Chapter 16 Solutions

College Physics

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