Example 1
I = 2 A, R = 10 Ω → P = 40 W.
Joule Power Loss Calculator
Type current I [A] and resistance R [Ω]. The calculator computes loss power P = I²·R. 2 A on 10 Ω is 40 W. In 60 s that is 2400 J of heat. Zero current gives zero loss.
Heat: Q = I²·R·t. Circuit power: P = U·I.
Result
Current I (A) and Resistance R (Ω). The result shows up here.
How it works
Power loss in a resistor is P = I² R. At 2 A and 10 Ω you get 40 W. At 1 A and 50 Ω you get 50 W. At 5 A and 0.5 Ω you get 12.5 W. I is squared: 4 A at 10 Ω is 160 W, not 80 W. At constant P, heat Q = P t. 40 W for 60 s is 2400 J.
The form has two fields: I in amperes and R in ohms. The result is in watts. A comma and a period are the same R: 0,5 and 0.5. This is loss in a resistor, not necessarily the power of a whole rail.
Typed 0 in I or R gives P = 0: no current or no resistance. Both fields need a number before 40 W can appear. A negative I is squared and gives the same P as |I|.
Q = I² R t is on the neighbouring page. Here power, there energy over time t. Q = P t. P = U I is power from voltage and current, not necessarily loss. For a purely thermal resistor they can be the same number.
R = ρ l / S lives on wire resistance. Here R is already an input. This calculator does not ask for cross-section or length.
Type 2 and 10, click Calculate, and match 40 W. The header symbol does not heat a resistor. Treat 40 W as the loss for that I and R pair, not as a nameplate.
How to use
- In the first field enter I in amperes, for example 2. Remember I goes into the formula squared.
- In the second field enter R in ohms, for example 10.
- Click Calculate. The calculator computes I² R. 2 A and 10 Ω give 40 W. Over 60 s that is 2400 J of heat.
- Typed 0 gives P = 0. Both fields need a number. Twice I is four times P.
- For Q = I² R t, open Joule heat. Power from U and I is on P = U·I.
Formula
P = I² · R
Energy over time t: Q = P · t = I² · R · t. Unit: W.
P from I²R loss
Watts from current squared: P = I²·R. 2 A and 10 Ω is 40 W. Over 60 s that is 2400 J of heat. I is squared.
- P
- Loss power [W]. 4 × 10 = 40 W. Not efficiency η.
- I
- Current [A]. Double I, four times P. 2 A in the example.
- R
- Resistance [Ω]. 10 Ω. No time in P. Time enters only in Q.
- Q
- Energy P·t. 40 W × 60 s = 2400 J. The same number as on the Joule heat card.
Real-life examples
Example 2
Small current, medium R: P = 50 W.
Example 3
Thick wire: P = 12.5 W.
Example 4
P = 25 W.
Example 5
LED: P = 0.06 W.
Example 6
P = 45 W.
Example 7
R = 0: P = 0 W.
Example 8
P = 10 W.
Example 9
P = 108 W.
Ways to use this calculator
- You compute 40 W from 2 A and 10 Ω.
- You compare loss with circuit power P = U I.
Frequently asked questions
What is P at 2 A and 10 Ω?
Loss power is 40 W, because 2² × 10 = 40. In 60 s the same loss is 2400 J of heat, since Q = P·t.
What about 5 A and 2 Ω?
You get 50 W. Five amperes squared is 25, times two ohms is fifty watts.
Which units do I type?
Current I [A], resistance R [Ω]. Result P [W], joules per second.
What if 0 A or 0 Ω?
Power is 0 W. No current or no resistance means no Joule loss. Both fields need a number.
How is this different from Q = I² R t?
Here you compute power, watts. There you compute energy over time t, joules. Q = P·t, so 40 W for 60 s is 2400 J.
How is this different from P = U I?
Here the loss in a resistor comes from I and R. There power comes from voltage and current. 2 A and 10 Ω is 40 W; 12 V and 0.5 A is 6 W on that card.
Is 4 A at 10 Ω equal to 80 W?
No. Current is squared, so 16 × 10 = 160 W, not 80. Doubling I quadruples the loss.
Does a negative I change P?
I² is positive. −2 A on 10 Ω is the same 40 W as +2 A, because the minus vanishes in the square.
Is 5 A and 0.5 Ω equal to 12.5 W?
Yes. 25 × 0.5 = 12.5. A fractional ohm is allowed.
Where is R = ρ l / S?
On the wire-resistance page. Here R is an input, not a result from length and area.
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.