   Chapter 16, Problem 50P

Chapter
Section
Textbook Problem

Two capacitors, C1 = 18.0 μF and C2 = 36.0 μF, are connected in series, and a 12.0-V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor. (b) Find the energy stored in each individual capacitor. Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances? (c) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy' as in part (a)? Which capacitor stores more energy in this situation, C1 or C2?

(a)

To determine
The equivalent capacitance and the energy stored in the capacitor.

Explanation

Given info: The capacitors C1=18.0μF and C2=36.0μF are connected in series. The potential difference applied by the battery is 12.0 V, The capacitors C1=18.0μF and C2=36.0μF are connected in series. The potential difference applied by the battery is 12.0 V.

Explanation:

Formula to calculate the equivalent capacitance is,

Ceq=C1C2C1+C2

Substitute 18.0μF for C1 and 36.0μF for C2 .

Ceq=(18.0μF)(36.0μF)(18.0μF)+(36.0μF)=12

(b)

To determine
The energy stored in each capacitor.

(c)

To determine
The potential difference.

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