Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=500μf.

Given that,.

R = 12 ohms

C = 500μf.

Time t = ? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp (-t/RC)

Where RC is the time constant

τ = RC = 12 * 500 * 10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 * Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp (-t/RC)

0.9999Qo = Qo•exp (-t / 0.006)

Divide both side by Qo

0.9999 = exp (-t / 0.006)

Take In of both sodes

In (0.9999) = In (exp (-t / 0.006))

-1 * 10^-4 = - t / 0.006

t = - 1 * 10^-4 * - 0.006

t = 6 * 10^-7 second

So it will take 6 * 10^-7 a for charge to reached 99.99% of it's maximum charge