Rate of heat generation in an electric circuit

AI Thread Summary
To find the rate of heat generation in a resistor, use the formula for power loss, which is P = I^2R. The current calculated at 2400 A appears incorrect, as it suggests an implausibly low resistance of 0.025 ohms for a 60-volt supply. Correct calculations show that with a 40-ohm resistor, the current should be 1.5 A. Therefore, the power loss in the resistor, and thus the heat generated, can be calculated using the correct current value. Accurate measurements of voltage and resistance are crucial for determining the heat generation rate.
benji
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How can I find the rate at which heat is generated in the resister in an electric circuit connected to a 40-ohm resistor that is embedded in a .20 kg solid substance in a calorimeter. The external portion of the circuit is connected to a 60-volt power supply. I calculated the current and it is 2400 A. Now I need to figure out what the "rate at which heat is generated in the resistor". I can figure out the rest of the problem once I have this information, but I don't know how to find it?

Thanks.
 
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2400A is a HUGE CURRENT...
You sure you got it right??

if you have the current passes through the resistor , the power loss in the resistor is just I^2R...
 
The only way to get 2400 amps out of 60 volts is if it is run across a resistance of .025 ohms. To get an accurate reading of the amperage from the power supply, attach the two points of a voltmeter to the power supply, set the voltmeter setting to 125V (I'm assuming DC) and check the ohms reading from the needle. My math gave me 1.5 amps when 60 volts is run through a 40 ohm resistor.

Here are the 3 equations derived from Ohms law in case they may help you out:

V = voltage
I = Current
R = Resistance

V = I x R
I = V/R
R = V/I

I = 60v/40 = 1.5 amps
R= 60/2400 = .025 ohms

I included the R calculation because I doubt your power supply is producing 60 volts with such incredibly reduced resistance.

Can't help you with the heat generation, I'm looking for that answer too, that's how how came across your thread. Good luck!
 
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