Answer:
a) RC = 1.03 mseg.
b) Qmax = CV = 85.2 μC
c) Q = 53.9 μC
Step-by-step explanation:
a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.
This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.
In this case , it can be calculated as follows:
ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.
b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:
Q = C V = 11.5 μF . 7.41 V = 85.2 μC
c) During the charging process, the charge increases following this equation:
Q = CV (1 - e⁻t/RC)
When t = RC, the expression for Q is as follows:
Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC