Example 1
I = 2 A, R = 10 Ω, t = 60 s → Q = 2400 J.
Joule Heat Calculator
Type I, R, and t. The calculator computes Q = I² R t. 2 A, 10 Ω and 60 s is 2400 J. Power loss for that I and R is 40 W. t cannot be negative.
Power loss: P = I²·R. Heat from mass: Q = c·m·ΔT.
Result
Current I (A), Resistance R (Ω) and Time t (s). The result shows up here.
How it works
Joule heat is the energy a resistor dumps when current runs through it: Q = I² R t. At 2 A, 10 Ω and 60 s you get 2400 J. That same I and R pair gives a loss P = I² R = 40 W, so Q is simply 40 W times 60 s. At 0.5 A, 100 Ω and 10 s you get 250 J.
The form has three fields in this order: current I in amperes, resistance R in ohms, time t in seconds. The result is in joules. A minute is 60, an hour is 3600. A comma and a period both parse: 0,5 and 0.5 are the same current.
I is squared, so twice the current is four times the heat, not two. 4 A at the same 10 Ω and 60 s is 9600 J, not 4800 J. R scales linearly: twice the resistance, twice Q, if I and t stay put.
Negative time has no meaning and the calculator will not run it. A typed zero in I, R, or t gives Q = 0: no current, no resistance, or no seconds elapsed. That is a deliberate zero, not a skipped field.
Power loss P = I² R is on the neighbouring page, without time. Here energy, there watts. Heat from mass Q = c m ΔT is another story: there you warm a body and measure a temperature jump, here you count joules from current in a resistor.
For a purely thermal resistor, electrical work W = U I t can be the same number as Q. On the work page you deliver energy to the circuit; here you ask how much of it sat down as heat in the resistor.
How to use
- In the first field enter current I in amperes. For example 2. Remember I goes into the formula squared.
- In the second field enter resistance R in ohms, for example 10. In the third, time t in seconds, for example 60. An hour is 3600, not 1.
- Click Calculate. The calculator computes I²·R·t. 2 A, 10 Ω and 60 s give 2400 J. For that I and R pair the loss is 40 W, so also 40 × 60.
- Negative time is rejected. Zero in any field gives Q = 0: no current, no resistance, or no time.
- Want watts without seconds? Open power loss P = I² R. Here it stays energy in joules.
Formula
Q = I² · R · t
Equivalently Q = P · t with P = I² · R. Unit: J.
Q, I, R, and t in Joule heat
Heat in a resistor: Q = I²·R·t. 2 A, 10 Ω and 60 s is 2400 J. At that I and R the loss power is 40 W.
- Q
- Heat in joules. 4 × 10 × 60 = 2400 J. Not charge Q = I t.
- I
- Current in amperes, squared. Double I and Q grows fourfold.
- R
- Resistance in ohms, 10 Ω in the example. Linear, not I².
- t
- Time in seconds. 60 s at 40 W is again 2400 J. Negative t is out.
- P
- Loss power P = I² R. 40 W. Q = P·t on this same card.
Real-life examples
Example 2
Small current, large R: Q = 250 J.
Example 3
Thick wire: Q = 1500 J.
Example 4
One hour: Q = 169200 J.
Example 5
Short pulse: Q = 1000 J.
Example 6
Small I, large R: Q = 6000 J.
Example 7
Q = 13500 J.
Example 8
R = 0: no losses, Q = 0 J.
Example 9
Q = 12000 J.
Example 10
LED with resistor: Q = 108 J.
Ways to use this calculator
- You compute 2400 J from 2 A, 10 Ω and 60 s.
- You check that 40 W for 60 s is the same 2400 J.
Frequently asked questions
How much Q at 2 A, 10 Ω and 60 s?
Q = 4 × 10 × 60 = 2400 J. P = I² R = 40 W, so also 40 × 60.
How much at 0.5 A, 100 Ω and 10 s?
Q = 0.25 × 100 × 10 = 250 J. That is I² R t with I [A], R [Ω], t [s].
Which units do I type?
I in amperes, R in ohms, t in seconds. Result in joules. An hour is 3600 s.
How is this different from P = I² R?
Here energy Q in joules. There power loss in watts, no time. Q = P t.
How is this different from Q = c m ΔT?
Here heat from current in a resistor. There from mass and a temperature change.
Why is twice I four times Q?
I is squared. 4 A at 10 Ω and 60 s is 9600 J, not 4800 J.
Does a comma in 0.5 work?
Yes. 0,5 and 0.5 mean the same I [A]. Then I² still enters the heat.
Where is W = U I t?
On electrical work. For a purely thermal resistor W and Q can be the same number.
Can I type t in minutes?
The field wants seconds. A minute is 60, ten minutes is 600.
What if R = 0?
Q = 0 J. An ideal lossless wire does not heat. One of the examples uses that card.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.