You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible internal resistance. find the time constant of the circuit. what is the charge of the capacitor 1.57 time constants after the circuit is closed? what is the charge after a long time?

1) The differential equation that models the RC circuit is:

(d/dt) V_capacitor + (V_capacitor/RC) = (V_source / RC)

Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (30.5 ohms) * (89.9-mf) = 2.742 s

2) C harge of the capacitor 1.57 time constants

1.57 * (2.742) = 4.3048 s

The solution of the differential equation is

V_capac (t) = (V_capac (0) - V_capac (∞)) e ^ (-t / T) + V_capac (∞)

Since the capacitor is initially uncharged V_capac (0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac (∞) = 9V

This means,

V_capac (t) = (-9V) e ^ (-t / T) + 9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V) e ^ (-4.3048 s / T) + 9V

V_capac (4.3048 s) = (-9V) e ^ (-4.3048 s / 2.742 s) + 9V

V_capac (4.3048 s) = (-9V) e ^ (-4.3048 s / 2.742 s) + 9V = - 1.87V + 9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s) = 89.9mF * (7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞) = 89.9mF * (9V) = 0.8091 C