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

the right answer is 0.3 meters per second