   Chapter 16, Problem 37P

Chapter
Section
Textbook Problem

Given a 2.50-μF capacitor, a 6.25-μF capacitor, and a 6.00-V battery, find the charge on each capacitor if you connect them (a) in series across the battery and (b) in parallel across the battery.

(a)

To determine
The charge on capacitance in series connection.

Explanation

Given info: The two capacitors C1=2.50μF and C2=6.25μF are connected in series. The potential difference applied by the battery is 6.00 V.

Explanation:

Formula to calculate the charge is,

Q=Ceq(ΔV)

• Ceq is the equivalent capacitance.
• ΔV is the potential difference.

The equivalent capacitance of two capacitors in series is given by,

Ceq=C1C2C1+C2

From the above equations,

Q=(C1C2C1+C2)(ΔV)

Substitute 2

(b)

To determine
The charge on capacitance in parallel connection.

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