Sounds like the resistor got to hot, indicating we exceeded its power rating. Typical resistors used in lab have a 250mW maximum power. To check this, we consider what happens at the moment the power supply is turned on. This is an instaneous event, so the capacitor maintains its original voltage. Just before the supply is turned on, the voltage across the capacitor is
Vc(0-) = 0 V
So just after the supply turns on, this voltage will remain. That is:
Vc(0+) = Vc(0-) = 0 V
That means the voltage across the resistor will be the full 5 volts of supply. That is, compared to the bottom wire as reference, the left side of the resistor is 5V and the right side of the resistor is 0V, for a full 5V across the resistor.
The power dissipated by the resistor is then V^2/R, where V is the voltage across the resistor (5V) and the resistance is 10 ohms. That is:
Pr = V^2 / R = 5^2 / 10 = 2.5W
This is 10 times the maximum power of a typical resistor, so it is no wonder it smoked!
We need to ensure the power dissipated by the resistor is less than 250mW, which means we need a larger resistor. In fact, we can compute the minimum resistance by solving the power equation above for R:
R = V^2 / Pr
R = 5^2 / .250 = 100
To be safe, we'll go larger than that, to say, 150 ohms.
With our new resistor, we should be safe from smoking any parts. Now the capacitor will charge from its initial voltage of 0V to eventually fully charged at 5V. That is, it will eventually look like an "open" circuit in steady-state. The time constant, tau, tells us how long it will take. The Thevenin resistance seen by the capacitor in this case is simply our resistor, so:
tau = RC = 150 ohms * 50mF = 7.5 seconds
If we wait for around 5 time constants, we should be quite close to the final value, so wait:
5 * tau = 5 * 7.5s = 37.5s
So around 40 seconds should do it.
Back to Transient Analysis Examples
(c) Steven H. VanderLeest, 2011