For the capacitor network shown in the figure, find the equivalent capacitance.(

Question

For the capacitor network shown in the figure, find the equivalent capacitance.(Figure 1)

Express your answer to two significant figures and include the appropriate units.

Part B

Find the charge on the capacitor a) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part C

Find the charge on the capacitor b) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part D

Find the charge on the capacitor c) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part E

Find the charge on the capacitor d) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part A

For the capacitor network shown in the figure, find the equivalent capacitance.(Figure 1)

Express your answer to two significant figures and include the appropriate units.

Part B

Find the charge on the capacitor a) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part C

Find the charge on the capacitor b) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part D

Find the charge on the capacitor c) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Part E

Find the charge on the capacitor d) when the entire combination is connected across a 20 V battery.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

A)
a and b are in parallel, their equivalent capacitance = a+b = 0.25 + 0.40 = 0.65 microF
c and d are in parallel, their equivalent capacitance = c+d = 0.35 + 0.50 = 0.85 microF
a+b and c+d are in series

Cnet = 0.65*0.85 / (0.65 + 0.85)
= 0.37 micro Farad

B)
Qnet = Cnet *20
= 0.37 micro Farad *20 V
= 7.4 micro Coulomb

This is charge on parallel combination of a and b
V across a and b = Q / (a+b)
= 7.4 micro Coulomb / 0.65 microF
=11.4 V

Qa = Va*Ca
= 11.4 V *0.25 micro F
= 2.8 micro coulomb

C)
Vb=Va = 11.4 V
Qb = Vb*Cb
= 11.4 V *0.40 micro F
= 4.6 micro coulomb

D)
V across c and d = 20 - 11.4 = 8.6 V
Qc = Vc*Cc
= 8.6 V *0.35 micro F
= 3.0 micro coulomb

I am allowed to answer only 4 parts at a time

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