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Question #43062

Difficulty: 8/10 | 🟢 Free

The Question

A uniform, purely resistive metallic rod of Young's modulus YY (in N/m2\text{N/m}^2) and coefficient of linear thermal expansion α\alpha (in K−1\text{K}^{-1}) is rigidly clamped between two fixed, perfectly insulating walls at an initial temperature T0T_0. The rod has a temperature-dependent electrical resistance R(T)=R0[1+β(T−T0)]R(T) = R_0[1 + \beta(T - T_0)], where R0R_0 is the resistance at T0T_0 (in Ω\Omega) and β\beta is the temperature coefficient of resistivity (in K−1\text{K}^{-1}). The rod is connected in parallel with a pure inductor, and this parallel combination is driven by an AC source.

In the steady state, the rod reaches a constant temperature T>T0T > T_0. At this steady state, the total RMS current drawn from the AC source is I0I_0 (in A\text{A}), and the overall power factor of the parallel circuit is cos⁡ϕ\cos\phi. The rod loses heat to the surroundings (which remain at T0T_0) at a rate K(T−T0)K(T - T_0), where KK is a positive constant (in W/K\text{W/K}).

Assuming the walls do not yield and the rod does not buckle, which of the following expressions gives the compressive thermal stress σ\sigma (in N/m2\text{N/m}^2) developed in the rod in the steady state?

Options

  • A.

    YαI02cos⁡2ϕR0K−I02cos⁡2ϕR0β\frac{Y \alpha I_0^2 \cos^2\phi R_0}{K - I_0^2 \cos^2\phi R_0 \beta}

  • B.

    YαI02R0K−I02R0β\frac{Y \alpha I_0^2 R_0}{K - I_0^2 R_0 \beta}

  • C.

    YαI02cos⁡ϕR0K−I02cos⁡ϕR0β\frac{Y \alpha I_0^2 \cos\phi R_0}{K - I_0^2 \cos\phi R_0 \beta}

  • D.

    YαI02cos⁡2ϕR0K+I02cos⁡2ϕR0β\frac{Y \alpha I_0^2 \cos^2\phi R_0}{K + I_0^2 \cos^2\phi R_0 \beta}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Wattless Current|First Law Application|Thermal Stress in Constrained Rods

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize AC circuit analysis with thermal expansion and steady-state heat transfer.

The primary trap lies in the misapplication of Joule heating in AC circuits.

Students often memorize P=Irms2RP = I_{rms}^2 R without analyzing branch currents.

A common mistake is using the total RMS current I0I_0 to calculate heat (P=I02RP = I_0^2 R), ignoring that in a parallel circuit, the current through the resistive rod is only the 'wattful' component I0cos⁡ϕI_0 \cos\phi.

Another trap is incorrectly writing power as I02Rcos⁡ϕI_0^2 R \cos\phi, confusing the general power formula P=VIcos⁡ϕP = V I \cos\phi with the I2RI^2 R form.

Furthermore, the problem requires setting up a self-consistent steady-state equation where the input power itself depends on the temperature via the temperature coefficient of resistivity β\beta, testing their ability to handle coupled electro-thermal systems.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptWattless Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree