   Chapter 16, Problem 41P

Chapter
Section
Textbook Problem

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.

(a)

To determine

The equivalent capacitance.

Explanation

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

Cpt=C6.00μFC3.00μFC6.00μF+C3.00μF

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

Cpb=C2.00μFC4.00μFC2.00μF+C4.00μF

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

Ceq=Cpt+Cpb

Therefore,

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

(b)

To determine

The charge on each capacitor.

(c)

To determine

The potential difference on each capacitor.

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