   Chapter 16, Problem 42P

Chapter
Section
Textbook Problem

Consider the combination of capacitors in Figure P16.42. (a) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance. (b) In diagram 1, find the equivalent capacitance of the three capacitors in parallel and redraw the diagram as a single battery and single capacitor in a loop. (c) Compute the charge on the single equivalent capacitor. (d) Returning to diagram 1, compute the charge on each individual capacitor. Does the sum agree with the value found in part (c)? (e) What is the charge on the 24.0-μF capacitor and on the 8.00-μF capacitor? Compute the voltage drop across (f) the 24.0-μF capacitor and (g) the 8.00-μF capacitor. Figure P16.42

(a)

To determine
The equivalent capacitance of series capacitors.

Explanation

Formula to calculate the equivalent capacitance of series capacitors is,

Cs=(24.0μF)(8.00μF)(24.0μF)+(8

(b)

To determine
The equivalent capacitance of parallel capacitors.

(c)

To determine
The total charge.

(d)

To determine
The charge on each capacitor.

(e)

To determine
The charge on 24.0μF capacitor.

(f)

To determine
The voltage drop across 24.0μF capacitor.

(g)

To determine
The voltage drop across 8.00μF capacitor.

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