Calculate Resistor Value & Power for Discharging Capacitor (1000uF, 300V)

In summary, the calculation shows that when using a 5k resistor, you will need 11W of power (2), but when using a 1k resistor, you only need only 9.78W of power (1). However, it is important to note that the capacitor doesn't discharge when the resistor is used, so always use a shorting stick to short the capacitor.
  • #1
Ozanai
2
0
Hi,

I’m trying to calculate the resistor value and power from discharging capacitor (1000uF, 300V).
Attached herewith is the calculation for discharge capacitor. the problem is that the result show that when I’m using 5k resistor I will need 11W (2), but when I’ll use 1k resistor, I only need only 9.78W (1). Is it possible?

Thanks,
Idan
 

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  • #2
Ozanai,

I’m trying to calculate the resistor value and power from discharging capacitor (1000uF, 300V).
Attached herewith is the calculation for discharge capacitor. the problem is that the result show that when I’m using 5k resistor I will need 11W (2), but when I’ll use 1k resistor, I only need only 9.78W (1). Is it possible?

Wow, you really got wrapped around the axle with your calculations, didn't you? Why are you involved with RMS values?

OK, take the 5k first. At 300 volts, the power dissipated in the resistor at the beginning is (300^2)/5000 = 18 watts. The time constant is 5000*1000E-6 = 5, so the voltage will drop to 300*(e^-1) = 110 volts in 5 seconds. At that time the power dissipated in the resistor will be (110^2)/5000 = 2.44 watts.

Next the 1k resistor. At 300 volts, the power dissipated in the resistor at the beginning is (300^2)/1000 = 90 watts. The time constant is 1000*1000E-6 = 1, so the voltage will drop to 300*(e^-1) = 110 volts in 1 second. At that time the power dissipated in the resistor will be (110^2)/1000 = 12.1 watts.

For each time constant period, the voltage drops to e^-1 = 0.37 of its previous value.

Ratch
 
  • #3
For single pulse application, calculate the joules stored in the capacitor, then calculate how much the joules will raise the conductive material in the resistor.
This is assuming all the joules go into heating the conductive material and none of the joules go into heating the porcelain or ceramic or whatever insulating materials is used.
Metal film resistor are generally considered the worst for pulse temperature rise, then thick film and then wirewound.
If you contact the wire wound resistor manufacture, they may be willing to tell you what the weight is and the material used for the wire.
Or you can always try a resistor and if it doesn't self destruct, it will probably work ok.
WORD OF WARNING: DON'T ASSUME THE RESISTOR DISCHARGES THE CAPACITOR. ALWAY USE A SHORTING STICK AND SHORT THE CAPACITOR.
We had a bad experience where a resistor open and someone got hurt.
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material, known as a dielectric.

2. How do you calculate the resistor value for discharging a capacitor?

The resistor value for discharging a capacitor can be calculated using the formula R = t/RC, where R is the resistance in ohms, t is the desired discharge time in seconds, and C is the capacitance in farads.

3. What is the power rating for discharging a 1000uF capacitor at 300V?

The power rating for discharging a capacitor can be calculated using the formula P = V^2/R, where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms. In this case, the power rating would be 90 watts.

4. Can I use any resistor to discharge a capacitor?

No, it is important to choose a resistor with the appropriate power rating to safely discharge a capacitor. Using a resistor with a lower power rating may cause it to overheat and potentially fail.

5. Why is it necessary to discharge a capacitor before handling it?

Capacitors can store a significant amount of electrical energy, which can be dangerous if released suddenly. Discharging a capacitor before handling it helps to prevent any potential electric shock.

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