How to Calculate Power (Electrical and Mechanical)

P = W / t

W here is work (or any energy change you are counting), in joules. t is time in seconds. P is in watts.

In a DC circuit, or for average values on many AC nameplates,

P = I V

I in amperes, V in volts, P still in watts, because 1 V·A = 1 W.

The power calculator on DevOkk is built around P = W/t: you enter work and time, with unit menus. No account. Electrical P = IV is the same watt, different inputs - do the product I×V yourself or treat IVΔt as the energy, then divide by t.

Watts are joules per second

A motor that does 2400 J of work in 8.0 s has

P = 2400 J / 8.0 s = 300 W

That is average power over those 8 seconds. If it did the same 2400 J in 2.0 s, P = 1200 W. Same energy, four times the power.

Minutes and hours. 2400 J in 8.0 min is 2400 / 480 = 5.0 W. Leaving t = 8 is the usual miss.

kWh is energy, not power. A kilowatt-hour is 3.6 × 10⁶ J. Utility bills charge energy. A 2.0 kW kettle run for 0.25 h uses 0.50 kWh. Calling the kettle “0.50 kW” because of that bill is backwards.

Horsepower. 300 W ÷ 746 ≈ 0.40 hp. Convert last. The power converter is that last step.

Electrical: P = IV and the resistor shortcuts

A resistor with 0.50 A through it and 24 V across it dissipates

P = (0.50 A)(24 V) = 12 W

In 10 s it dumps 120 J as heat (P × t). If you only have I and R = 48 Ω, P = I²R = (0.25)(48) = 12 W. If you only have V and R, P = V²/R = 576/48 = 12 W.

AC caveat. For AC, the wall-socket “120 V” is RMS, and P_avg = I_rms V_rms cosφ for a linear load. Using peak × peak overstates by a factor of two if you forget RMS. Intro problems that say “12 V DC” do not have that issue.

Series and parallel. Total power is the sum of the powers in each element (energy is conserved). You cannot always use the source voltage times one branch current and call it the whole circuit without looking at the diagram.

A motor doing 2400 J in 8.0 s

A winch lifts a load. The gain in gravitational PE is mgh = 2400 J (you already computed PE). The lift takes 8.0 s.

P_avg = 2400 / 8.0 = 300 W

If the motor is 60% efficient, the electrical input power is 300 / 0.60 = 500 W. At 24 V DC that would be I = P/V = 500/24 ≈ 21 A. Efficiency is why nameplate watts and “useful” mechanical watts differ. The calculator’s 300 W is the mechanical side if 2400 J was the useful work.

A second check with force and speed: if the cable tension is 400 N and the load rises at a steady 0.75 m/s, instantaneous P = Fv = 300 W. Same number, different measurements. If those two lab methods disagree, one of F, v, m, h, or t is wrong - not the definition of a watt.

When P = W/t or P = IV is the wrong first equation

Energy without time. 116 J of kinetic energy is not a power until you say how quickly it was delivered. A 116 J impact in 2.0 ms is 58 kW average during the crumple; that is why collisions are violent even when the energy looks modest. See kinetic energy.

Torque and rpm. Rotating machinery uses P = τω with ω in rad/s. Converting rpm as if it were rad/s (forgetting 2π/60) is the mechanical analog of leaving minutes in t.

dB, VA, and var. Apparent power in AC is VI (volt-amperes). Reactive power is not watts. A “500 VA” supply is not automatically 500 W.

Human output. “A person does 200 J of work” on a stair problem still needs a time to become watts. Do not paste 200 into a power field.

Unit traps

Mixing kJ with seconds and labeling the result kW without dividing by 1000 correctly: 2.4 kJ / 8.0 s = 300 W = 0.30 kW. Writing 2.4/8 = 0.30 kW happens to work; writing 2400 kJ / 8 s does not.

BTU/h is a power unit (energy per time). BTU alone is energy. The converter can switch BTU/h and watts; it cannot rescue a missing hour.

Compute in watts, convert once

Enter work and time in the power calculator after they share a coherent system (J and s, or kJ and s with the tool’s units). For P = IV, multiply current and voltage, then treat that as watts. Use the power converter for hp or kW display. A fuller unit table is in energy and power units.