In the figure , C1 = C5 = 8.0 F and C2= C3 = C4 = 4.8 F . The applied potential
ID: 3163375 • Letter: I
Question
In the figure , C1 = C5 = 8.0 F and C2= C3 = C4 = 4.8 F . The applied potential is Vab = 230 V .
Part A What is the equivalent capacitance of the network between points a and b? Express your answer using two significant figures.
Part B Calculate the charge on capacitor C1. Express your answer using two significant figures.
Part C Calculate the potential difference across capacitor C1. Express your answer using two significant figures.
Part D Calculate the charge on capacitor C2. Express your answer using two significant figures.
Part E Calculate the potential difference across capacitor C2. Express your answer using two significant figures.
Part F Calculate the charge on capacitor C3. Express your answer using two significant figures.
Part G Calculate the potential difference across capacitor C3. Express your answer using two significant figures
Part H Calculate the charge on capacitor C4. Express your answer using two significant figures.
Part I Calculate the potential difference across capacitor C4. Express your answer using two significant figures.
Part J Calculate the charge on capacitor C5. Express your answer using two significant figures.
Part K Calculate the potential difference across capacitor C5. Express your answer using two significant figures.
C C. C. C be 3 2 aExplanation / Answer
Here ,
for the equivalent capacitor
first C3 and C4 are in series
1/C6 = 1/4.8 + 1/4.8
C6 = 2.4 uF
Now, for C2 and C6 in parallel
C7 = 2.4 + 4.8 = 7.2 uF
for C1 , C5 and C7 in parallel
1/Ceq = 7.2 + 1/8 + 1/8
Ceq = 2.57 uF
the equivalent capacitance of the network between points a and b is 2.57 uF
part B)
charge on capacitor C1 = V * Ceq
charge on capacitor C1 = 230 * 2.57 uC
charge on capacitor C1 = 591.4 uC
part C)
potential across C1 = 591.4/8 V
potential across C1 = 73.9 V
part D)
charge on capacitor 2 = 591.4 * 4.8/(2.4 + 4.8)
charge on capacitor 2 = 394.3 uC
the charge on capacitor 2 is 394.3 uC
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