Find the charge Q(t) on the capacitor?

Question

A battery (9.284 V) is connected to a resistor (9856 ohms) and a capacitor (15.73 mcF). Calculate the charge Q(t) on the capacitor if initially the stored charge was zero. Show your work please, if you want to be chosen a best answer.

The diagram can be found in the following link:

http://www.geocities.com/trunks11111/charge.jpg

Answer 1

Let ξ=9.284V, R = 9856 Ω, C = 15.73 μF (I am interpreting your mcF to mean microfarads).

The Kirchhoff loop rule states that the voltage drop across each circuit element in a closed circuit sums to zero:

ξ - IR - q/C = 0

Substitute I = dq/dt and multiply both sides of equation by C.

Cξ - RCdq/dt - q = 0

Rearrange so that all q terms are on one side of equation, and integrate.

1/(RC)∫dt = ∫dq/(Cξ - q)

Time is integrated from t = 0 to t = t, and charge is integrated from q = 0 to q = Q.

t/(RC) = - ln [(Cξ - Q)/Cξ]

Both sides are raised to an exponent using the base e, and rearranged to form:

Q(t) = Cξ(1 - exp( - t / RC))

You will often see the substitution τ = RC, which is called the RC time constant.

Plugging in your numbers:

Q(t) = (15.73E-6*9.284)*(1 - exp( - t / (9856*15.73E-6))

Q(t) = (1.460E-4 C)*(1 - exp(-6.450t))

Answer 2

Q(t)=(9.284V)(1-e^(-t/(9856*15.73*10^-6)))
Q(t)=9.284*(1-e^(-t/0.155))

because charging capacitors follow the relation:

Q(t)=Qfinal*(1-e^(-t/RC)) when in a series setup as shown.

Answer 3

Q(t) = V ( 1 - exp (-t/RC))