The symbol t in the capacitor discharge formula q= (q0e)^(–t/RC)

[hidemi]

Homework Statement

In the capacitor discharge formula q= (q0e)^(–t/RC) the symbol t represents:
A)the time constant
B)the time it takes for C to lose the fraction 1/e of its initial charge
C)the time it takes for C to lose the fraction (1 – 1/e) of its initial charge
D)the time it takes for C to lose essentially all of its initial charge
E)none of the above
The answer is E.

Relevant Equations

q= (q0e)^(–t/RC)

If none of the above is correct, what is a good definition of the symbol t (time)?

 

[kuruman]

Clock time. It's the same $t$ as in $x=v_0t+\frac{1}{2}at^2$. Here the imaginary clock starts ticking at $t=0$ when the capacitor starts discharging.

 

[hidemi]

Clock time. It's the same $t$ as in $x=v_0t+\frac{1}{2}at^2$. Here the imaginary clock starts ticking at $t=0$ when the capacitor starts discharging.

Ok I see.
Another question is that why we need to divide the time by the time constant (tao)?

 

[kuruman]

Ok I see.
Another question is that why we need to divide the time by the time constant (tao)?

As you know, the charge on the capacitor at any time $t$ is given by $q(t)=q_0e^{-\frac{t}{RC}}.$ The product $RC$ has dimensions of time. So we define time constant $\tau=RC$ and substitute in the equation to get $q(t)=q_0e^{-\frac{t}{\tau}}.$

 

[hidemi]

As you know, the charge on the capacitor at any time $t$ is given by $q(t)=q_0e^{-\frac{t}{RC}}.$ The product $RC$ has dimensions of time. So we define time constant $\tau=RC$ and substitute in the equation to get $q(t)=q_0e^{-\frac{t}{\tau}}.$

Ok I see. Thank you.

 

[berkeman]

Another helpful mental picture is to understand that for each time constant $\tau$ the level of charge goes down to $\frac{1}{e}$ of its initial level Q. That is about 37% left after each time constant. So after one time constant Q goes to Q*0.37Q, after 2 time constants, Q has gone down to Q*(0.37)^2, and so on.

https://www.eecs.tufts.edu/~dsculley/tutorial/rc/dischargeCurve.jpg