Assertion–Reason (electrical power and heating): Assertion (A) Power loss in a conductor is P = I^2 R. Reason (R) With current density J under an applied electric field E, the heat developed per unit volume per second equals J·E.

Difficulty: Easy

Correct Answer: Both A and R are true and R is correct explanation of A

Explanation:


Introduction / Context:
Joule heating describes electrical power converted to heat in conductors and resistive media. The macroscopic formula P = I^2 R is widely used, while the microscopic expression for volumetric heating q̇_v = J·E connects material properties and field distributions to thermal power density.


Given Data / Assumptions:

  • Ohmic conductor with J = σE (linear isotropic medium).
  • Steady sinusoidal or DC conditions (instantaneous relations apply to both).
  • Uniform cross-section A and length ℓ for simple derivation.


Concept / Approach:

The local power density is q̇_v = J·E. Integrating over the conductor volume gives total power. Using Ohm’s law relationships (V = IR, I = JA, E = V/ℓ), we can reduce the volume integral to the lumped formula P = I^2 R, demonstrating that the microscopic statement explains the macroscopic result.


Step-by-Step Solution:

Start with q̇_v = J·E = σE^2 = J^2/σ.Total power P = ∫_V q̇_v dV = ∫_V J·E dV.For uniform conductor: J = I/A, E = V/ℓ, so P = (I/A)(V/ℓ) Aℓ = IV.Using V = IR gives P = I^2 R, proving A and showing R explains A.


Verification / Alternative check:

The same conclusion follows from P = V^2/R by substituting I = V/R, consistent with energy conservation and circuit theory.


Why Other Options Are Wrong:

  • If R were false, the derivation from fields to circuit formula would fail, but it is standard.


Common Pitfalls:

  • Confusing magnitude J·E with vector product sign conventions; in passive media J is parallel to E, so J·E ≥ 0.


Final Answer:

Both A and R are true and R is correct explanation of A

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