A parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is connected in series to a 4.89 μF capacitor. This three‑capacitor combination is connected to a 15.5 V battery. Determine the charge on each capacitor.

a. i. 35.96 μC  b. i. 11.98 μC ii. 24.04 μC

Explanation:

We need to find the total capacitance of the system C.

The total capacitance of the parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is C' = 1.47 μF + 2.95 μF = 4.42 μF.

C' = 4.42 μF is in series with the 4.89 μF capacitor and for a series combination of capacitors, we have the total capacitance, C from

1/C = 1/4.42 μF + 1/4.89 μF

1/C = (4.42 μF + 4.89 μF)/(4.42 μF × 4.89 μF)

1/C = 9.31 μF/21.6138 μF²

C = 21.6138/9.31 μF

C = 2.32 μF

So, the total charge in the circuit Q = CV where C = total capacitance = 2.32 μF and v = voltage = 15.5 V

So, Q = CV

Q = 2.32 μF × 15.5 V

Q = 35.96 μC

i. The charge on the 4.89 μF capacitor

Since the 4.89 μF is in series with C', the total charge flowing i the circuit is the total charge in the 4.89 μF capacitor. So, its charge Q = 35.96 μC

b. The charge in the 1.47 μF and 2.95 μF capacitors.

To find the charge in the 4.89 μF and 2.95 μF capacitors, we need to find the voltage across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor. The voltage, V' across the 4.89 μF capacitor, since Q = CV', V' = Q/C = 35.96 μC/4.89 μF = 7.35 V

So, the voltage V" across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor, C' is V" = 15.5 V - V' (since V' + V" = 15.5 V).

So, V" = 15.5 V - V'

V" = 15.5 V - 7.35 V

V" = 8.15 V

i. The charge on the 1.47 μF capacitor

Using Q' = CV" where Q' = charge across capacitor, C = 1.47 μF and V" = 8.15 V.

So, Q' = CV"

Q' = 1.47 μF × 8.15 V

Q' = 11.98 μC

ii. The charge on the 2.95 μF capacitor

Using Q" = CV" where Q' = charge across capacitor, C = 2.95 μF and V" = 8.15 V.

So, Q" = CV"

Q" = 2.95 μF × 8.15 V

Q" = 24.04 μC

answer;

d. are made mostly of liquid

explanation;

-the general   composition and structure planets includes; they have a thick gas and liquid layer covering a small,solid core. mostly the gases are hydrogen and helium atmosphere and layers of liquid hydrogen. for instance neptune and uranus are examples of the outer planets, they have a core covered by ice and ammonia.

-neptune's atmosphere is hydrogen and helium with a bit of methane so there is no oxygen to breath and the interior is made up of frozen water and ammonia with a rock and iron core. thus, there is no liquid water or solid surface to survive.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2