In the figure (Figure 1), each capacitance C1C1 is 6.9 μFμF , and each capacitance C2C2 is 4.6 μFμF .
A:
Compute the equivalent capacitance of the network between points a and b.
Express your answer using two significant figures.
|
Ceq = |
??? |
B:
Compute the charge on the capacitor C1C1 nearest to a when VabVab = 430 VV .
Express your answer using two significant figures.
Qa1 = ????
C:
Compute the charge on the capacitor C1C1 nearest to b when VabVab = 430 VV .
Qb1 =???
D:
Compute the charge on the capacitor C2C2 nearest to a and b when VabVab = 430 VV .
Express your answer using two significant figures.
Q2 =???
E:
With 430 VV across a and b, compute VcdVcd.
Express your answer using two significant figures.
Vcd =????
Homework Answers
solving the circuit from right to left
last 3 capacitors ar ein series
resultant of these capacitors is = 6.9 μF/3 = 2.3 μF
this is parallel with c2 = 2.3+4.6 = 6.9 μF
again c1.6.9μF and c1 are in series = 6.9/3 = 2.3 μF
now 2.3μF and c2 are in parallel = 2.3+4.6 = 6.9μF
near a and b
c tot = 6.9/3 = 2.3 μF
total charge q tot = 2.3*10^-6*430 = 0.989*10^-3 C
b)
the charge on the capacitor C1 nearest to a = 0.989*10^-3 C
because in a sereis connection change is same on the capacitor
c) the charge on the capacitor C1 nearest to b = 0.989*10^-3 C
d) charge on c2 between a and b
voltage across c2 = 0.989*10^-3/6.9*10^-6 = 143.3 V
charge on c2 between a and b = 4.6*10^-6*143.3 = 0.659*10^-3 C
e) vcd = 47.77 V
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