# Capacitor and Variable resistor

1. Apr 19, 2005

### 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

2. Apr 19, 2005

### CartoonKid

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

3. Apr 19, 2005

### 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?

4. Apr 19, 2005

### dtang0

the way it is worded, cartoon kid is correct.

jw, but what was your reasoning behind this?

Last edited: Apr 19, 2005
5. Apr 19, 2005

### CartoonKid

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.

6. Apr 19, 2005

### 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.