Joule''s Law

Calculate heat generated by electric current.

Inputs

I

R

t

Formula Interpretation

Joule''s Law

Q = I^2 R t

$Q$ (J) is the heat generated in time $t$ (s); it is proportional to $I$ ² (A²), $R$ (Ω), and $t$ . The quadratic dependence on current is the most critical aspect: doubling current quadruples heat. The formula assumes constant resistance; if resistance changes with temperature (e.g. PTC thermistors), the heat must be computed by integration over time.

Solve for Current

I = \sqrt{\dfrac{Q}{Rt}}

Solve for the current needed to generate heat $Q$ (J) in resistance $R$ (Ω) over time $t$ (s). Used in electric-heater design: compute the required working current from power demand and resistance, then select the appropriate conductor and switching devices.

Solve for Resistance

R = \dfrac{Q}{I^2 t}

Solve for equivalent resistance when heat $Q$ (J), current $I$ (A), and time $t$ (s) are known. Back-calculating resistance from measured power dissipation and current helps in thermal modelling and fault location.

Solve for Time

t = \dfrac{Q}{I^2 R}

Solve for the time needed to generate target heat $Q$ (J) at current $I$ (A) through resistance $R$ (Ω). Directly applicable to estimating heating times for electric water heaters, industrial dryers, and reflow-soldering ovens.

Knowledge Points

Physical origin of Joule heating

When current flows through a conductor, drifting electrons collide with lattice ions, converting kinetic energy into lattice vibration (heat). The instantaneous power dissipated is $P = I^2 R$ (W) and total heat is $Q = Pt = I^2 Rt$ (J). Electric heating appliances (heaters, dryers, electric stoves) deliberately exploit this effect; in conductors, transformers, and motors, Joule heat is a parasitic loss to be minimised through proper sizing and cooling.

Quadratic current relationship and safety design

Heat scales as $I^2$ : a 20% overload ( $I$ = 1.2× rated) produces 44% excess heat; a 10× fault current generates 100× normal heat in the same time. This explains why overcurrent protection devices must respond progressively faster as fault current increases. Fuses and circuit breakers are characterised by their $I^2t$ trip curves, which define the maximum energy let-through before they open. Even a modest sustained overload, if undetected, gradually degrades insulation and shortens equipment life.

Thermal energy accumulation and steady-state temperature rise

At constant current with fixed heat dissipation, conductor temperature rises asymptotically toward a steady-state value $\Delta T_{\text{ss}} = P/(hA)$ , where $h$ is the combined convective–conductive heat transfer coefficient and $A$ is the heat-dissipation area. The time constant $\tau_{\text{th}} = mc_p/(hA)$ governs how quickly the temperature approaches steady state. Equipment rated for continuous duty is designed to the steady-state temperature; intermittently loaded equipment can tolerate higher currents if the duty cycle keeps the average heat within thermal limits.

Thermal time constant and transient thermal design

The thermal time constant $\tau_{\text{th}} = mc_p / (hA)$ (s) quantifies the thermal inertia of a conductor or component, where $m$ is mass and $c_p$ is specific heat capacity. Small, light conductors (thin PCB traces) have very short thermal time constants (milliseconds to seconds) and heat up almost instantly with overcurrent. Large transformer windings and motor coils have time constants of minutes to tens of minutes, tolerating brief overloads without exceeding temperature limits. Knowing $\tau_{\text{th}}$ is essential for setting overcurrent trip delay curves and verifying short-circuit thermal withstand ratings.

Example

A heating wire has $R = 5\,\Omega$ , carries $I = 2\,\text{A}$ , and is energized for $t = 10\,\text{s}$ . Find the heat $Q = 200\,\text{J}$ and convert to kJ.

Step 1 — Compute heat

Q = I^2 R t = 2^2 \times 5 \times 10 = 200\,\text{J}

Step 2 — Convert to kJ

200\,\text{J} = 0.2\,\text{kJ}

Therefore the heat is $Q = 200\,\text{J}$ , which equals $0.2\,\text{kJ}$ .

Extended Knowledge

•PCB trace current capacity and IPC-2221：PCB copper trace heating power is $P = I^2 R_{\text{trace}}$ , where $R_{\text{trace}} = \rho L / (W \cdot t)$ . Temperature rise depends on trace width, copper weight, layer position (external traces dissipate heat better than internal ones), and PCB thermal conductivity. IPC-2221 (superseded in detail by IPC-2152) provides current-vs-width curves for external and internal layers at allowable temperature rises of 10–30°C. Exceeding these limits causes solder-mask blistering, delamination, or trace burnout.
•AC heating and harmonic currents：For AC circuits, Joule heating uses the RMS current: $Q = I_{\text{rms}}^2 R t$ . Non-sinusoidal currents (switch-mode power supplies, VFDs) contain harmonics whose total RMS is $I_{\text{rms}}^2 = I_1^2 + I_3^2 + I_5^2 + \cdots$ . Harmonic currents cause extra heating in transformers and neutral conductors beyond what is expected from fundamental-frequency calculations alone. Distribution transformers and cables serving harmonic-rich loads must be derated accordingly, typically following IEEE 519 guidelines.
•Thermal runaway and positive-temperature-coefficient effects：Thermal runaway occurs when rising temperature increases power dissipation in a positive-feedback loop: higher $T$ → higher $R$ (PTC) or lower $R$ (NTC) → more heat. In lithium-ion batteries, thermal runaway from overcharge or external heat can trigger electrolyte decomposition, gas generation, and fire. In power semiconductors, negative temperature coefficient characteristics can cause current hogging between parallel devices. Design countermeasures include current-sharing resistors, thermal fuses, NTC-based current limiters, and rigorous thermal simulation.
•Fuse $I^2t$ characteristics and protection coordination：A fuse''s melting energy is characterised by its pre-arcing $I^2t$ (ampere-squared-seconds). Fast-acting semiconductor fuses have very low $I^2t$ values and open within sub-millisecond intervals to protect IGBTs and thyristors. Time-delay (gG-type) fuses tolerate several times rated current for seconds, allowing motors and transformers to start without nuisance tripping. Coordinating upstream and downstream protective devices requires that the downstream device''s maximum clearing $I^2t$ is less than the upstream device''s minimum melting $I^2t$ , ensuring selective tripping that confines outages to the faulted zone.

Joule''s Law

Calculate heat generated by electric current.

Inputs

I

R

t

Formula Interpretation

Joule''s Law

Q = I^2 R t

Solve for Current

I = \sqrt{\dfrac{Q}{Rt}}

Solve for Resistance

R = \dfrac{Q}{I^2 t}

Solve for Time

t = \dfrac{Q}{I^2 R}

Knowledge Points

Physical origin of Joule heating

Quadratic current relationship and safety design

Thermal energy accumulation and steady-state temperature rise

Thermal time constant and transient thermal design

Example

A heating wire has $R = 5\,\Omega$ , carries $I = 2\,\text{A}$ , and is energized for $t = 10\,\text{s}$ . Find the heat $Q = 200\,\text{J}$ and convert to kJ.

Step 1 — Compute heat

Q = I^2 R t = 2^2 \times 5 \times 10 = 200\,\text{J}

Step 2 — Convert to kJ

200\,\text{J} = 0.2\,\text{kJ}

Therefore the heat is $Q = 200\,\text{J}$ , which equals $0.2\,\text{kJ}$ .

Extended Knowledge

•PCB trace current capacity and IPC-2221：PCB copper trace heating power is $P = I^2 R_{\text{trace}}$ , where $R_{\text{trace}} = \rho L / (W \cdot t)$ . Temperature rise depends on trace width, copper weight, layer position (external traces dissipate heat better than internal ones), and PCB thermal conductivity. IPC-2221 (superseded in detail by IPC-2152) provides current-vs-width curves for external and internal layers at allowable temperature rises of 10–30°C. Exceeding these limits causes solder-mask blistering, delamination, or trace burnout.
•AC heating and harmonic currents：For AC circuits, Joule heating uses the RMS current: $Q = I_{\text{rms}}^2 R t$ . Non-sinusoidal currents (switch-mode power supplies, VFDs) contain harmonics whose total RMS is $I_{\text{rms}}^2 = I_1^2 + I_3^2 + I_5^2 + \cdots$ . Harmonic currents cause extra heating in transformers and neutral conductors beyond what is expected from fundamental-frequency calculations alone. Distribution transformers and cables serving harmonic-rich loads must be derated accordingly, typically following IEEE 519 guidelines.
•Thermal runaway and positive-temperature-coefficient effects：Thermal runaway occurs when rising temperature increases power dissipation in a positive-feedback loop: higher $T$ → higher $R$ (PTC) or lower $R$ (NTC) → more heat. In lithium-ion batteries, thermal runaway from overcharge or external heat can trigger electrolyte decomposition, gas generation, and fire. In power semiconductors, negative temperature coefficient characteristics can cause current hogging between parallel devices. Design countermeasures include current-sharing resistors, thermal fuses, NTC-based current limiters, and rigorous thermal simulation.
•Fuse $I^2t$ characteristics and protection coordination：A fuse''s melting energy is characterised by its pre-arcing $I^2t$ (ampere-squared-seconds). Fast-acting semiconductor fuses have very low $I^2t$ values and open within sub-millisecond intervals to protect IGBTs and thyristors. Time-delay (gG-type) fuses tolerate several times rated current for seconds, allowing motors and transformers to start without nuisance tripping. Coordinating upstream and downstream protective devices requires that the downstream device''s maximum clearing $I^2t$ is less than the upstream device''s minimum melting $I^2t$ , ensuring selective tripping that confines outages to the faulted zone.