Solution Found!
Solved: In Fig. P24.53, C1 = C5 = 8.4 µF and C2 = C3 = C4
Chapter 24, Problem 57P(choose chapter or problem)
In Fig. P24.57, \(C_{1}=C_{3}=8.4 \mu F\) and \(C_{2}=C_{3}=C_{4}=4.2 \mu F\). The applied potential is \(V_{a b}=220 V\). (a) What is the equivalent capacitance of the network between points and
(b) Calculate the charge on each capacitor and the potential difference across each capacitor.
Equation transcription:
Text transcription:
C{1}=C{3}=8.4 \mu F
C{2}=C{3}=C{4}=4.2 \mu F
V{a b}=220 V
Questions & Answers
QUESTION:
In Fig. P24.57, \(C_{1}=C_{3}=8.4 \mu F\) and \(C_{2}=C_{3}=C_{4}=4.2 \mu F\). The applied potential is \(V_{a b}=220 V\). (a) What is the equivalent capacitance of the network between points and
(b) Calculate the charge on each capacitor and the potential difference across each capacitor.
Equation transcription:
Text transcription:
C{1}=C{3}=8.4 \mu F
C{2}=C{3}=C{4}=4.2 \mu F
V{a b}=220 V
ANSWER:Solution 57P
Step 1
The following diagram shows the simplified circuitry.
