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

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SUMMARY

This discussion focuses on calculating the resistor value and power required for discharging a 1000uF, 300V capacitor. The calculations reveal that using a 5kΩ resistor results in an initial power dissipation of 18W, while a 1kΩ resistor leads to an initial power dissipation of 90W. The time constants for the respective resistors are 5 seconds and 1 second, with power dissipation decreasing over time. A critical safety warning emphasizes the necessity of using a shorting stick to safely discharge capacitors.

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  • Understanding of capacitor discharge principles
  • Knowledge of Ohm's Law and power calculations
  • Familiarity with time constant calculations in RC circuits
  • Awareness of resistor types and their thermal characteristics
NEXT STEPS
  • Study the calculations for capacitor discharge using different resistor values
  • Learn about the thermal characteristics of various resistor types, including metal film and wirewound
  • Research safety protocols for discharging capacitors in electronic circuits
  • Explore the concept of joules stored in capacitors and their impact on resistor heating
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Electronics engineers, hobbyists working with capacitors, and anyone involved in designing circuits that require safe capacitor discharge methods.

Ozanai
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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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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
 
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.
 

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