Mechanism of power dissipation in a resistor In practical circuits, by what physical process is electrical power dissipated inside a resistor?

Introduction / Context:
Resistors convert electrical energy into another form as current flows through them. Understanding what that form is explains why resistors warm up, require power ratings, and sometimes need heat sinks or derating with temperature.

Given Data / Assumptions:

• Ohmic conductor under steady DC or AC conditions.
• No special thermoelectric or radiative devices involved.
• Macroscopic view of energy conversion.

Concept / Approach:
Electrical power in a resistor is P = V * I = I^2 * R = V^2 / R. On the microscopic level, moving electrons collide with the lattice, transferring energy and increasing lattice vibrations. This manifests as temperature rise—thermal energy (heat). The resistor then dissipates this heat to its surroundings by conduction, convection, and radiation.

Step-by-Step Solution:
Compute electrical power using P = I^2 * R or P = V^2 / R.Recognize conversion: electrical energy → thermal energy inside the resistive material.Observe: body temperature of the resistor increases as it dissipates heat.Thermal equilibrium: heat output balances ambient heat removal, defining allowable continuous power (the power rating).

Verification / Alternative check:
Touch-safe observation and IR thermography show resistors warming under load. Datasheets specify temperature rise versus power and ambient conditions.

Why Other Options Are Wrong:
By resistance/voltage/current: these are parameters in the electrical equations, not mechanisms of energy conversion. The physical manifestation is heat.

Common Pitfalls:
Ignoring power ratings and placing low-wattage resistors in high-dissipation positions, leading to overheating or failure.

Final Answer:
by heat