Capacitor and Variable resistor

[coldturkey]

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.220\mu F capacitor. The capacitor is charged to 5.00V, then discharged through the resistor. The time for the potential difference across the plates to decrease to 0.800V is measured by a clock inside the game. If the range of discharge times that can be handled effectivly is from 10.0\mu s to 6.00ms, what should be the resistance range of the resistor?

I have solved the problem and I get a maximum resistance of 27272.7\Omega and a minimum resistance of 45.45\Omega.
But these values seem a bit too large.

The way I did it:
I = q/t
q = CV
so I = CV/t
and R = V/I

and solved it for all 4 cases:
(max voltage, largest discharge time)
(max voltage, smallest discharge time)
(min voltage, largest discharge time)
(min voltage, smallest discharge time)

An found there are two different values for the resistor:
27272.7\Omega and 45.45\Omega.

Does anyone know if there is anything I have done wrong?
Many thanks

 

[CartoonKid]

I think you can use the formula:
V=V_{0}e^{-\frac{t}{RC}}

 

[coldturkey]

I tried it using your formula and I get 24.8 ohms and 14882 ohms.
This is roughly half of what I got before.
Any ideas?

 

[dtang0]

the way it is worded, cartoon kid is correct.

jw, but what was your reasoning behind this?

and solved it for all 4 cases:
(max voltage, largest discharge time)
(max voltage, smallest discharge time)
(min voltage, largest discharge time)
(min voltage, smallest discharge time)

 

[CartoonKid]

I tried it using your formula and I get 24.8 ohms and 14882 ohms.
This is roughly half of what I got before.
Any ideas?

In a RC circuit, when a capacitor is discharging, the charges, current and voltage across the capacitor are decreasing exponentially. It's a continuous process. The bigger the R, the slower the discharing process.

 

[coldturkey]

well I wasnt sure what the relationships were all about so I just decided to try all possible cases and see what I came up with.