A circuit with a capacitor took .138s to discharge from its max charge Qmax to half of that max charge once the battery was removed. In this series circuit there also is one resistor with 100 G(ohm symbol). The voltage drop is 12V across the battery.
a)How much time would it take the capacitor to discharge from its max charge to 1/16th of that max charge?
1/2Qmax => t=1.38s
1/4Qmax => 1.38/(.5)= .276s
1/6Qmax => .276s/(.25) = 1.104 s
b)What is the time constant of this circuit?
I know that time constant = RC but we are given two unknowns the time constant and the capacitance. R would be 100 x 10^9 or 1 x 10^11.
Find the time constant RC first.
1/2=e^(-t/RC) solve for RC
Then,
1/16=e^-t/RC, solve for time for a).
To find the time constant of the circuit, you need to determine the capacitance (C) of the capacitor.
The time constant (τ) of a RC circuit is given by the product of the resistance (R) and the capacitance (C), i.e., τ = RC.
In this case, you are given the resistance (R) as 100 GΩ (which is equivalent to 1 x 10^11 Ω). However, the capacitance (C) is not provided.
To determine the time constant, we need an additional piece of information.