Answer to Question #84095 in Electric Circuits for Oliver Jones

Question

An example Capacitor is 24v and 1500μF and has a time constant of 0.15 Seconds.
I need to determine (using a discharge equation) the total charge, in coulombs discharged in the first 5 time constants and then predict how long it would take to discharge to 0.03 coulombs.

Answer 1

The discharge equation:

$$V(t)=V_0e^{-t/T},$$

where

$$T=RC$$

is a time constant. The total charge in the first 5 time constants thus is:

$$Q=C\cdot V(5T)=C\cdot V_0e^{-5}=$$ $$=1.5\cdot10^{-3}\cdot24\cdot e^{-5}=2.43\cdot10^{-4} \textrm{ C}.$$

Now predict how long it would take to discharge our capacitor to 0.03 coulombs (when the voltage of course will be

$$V=Q/C=0.03/(1.5\cdot10^{-3})=20 \textrm{ V}$$

). Express

$$t$$

from the first formula:

$$t=-RC\cdot {\textrm{ln}\frac{V}{V_0}}=-0.15\cdot {\textrm{ln}\frac{20}{24}}=0.027\textrm{ s}$$