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Question #43059

Difficulty: 8/10 | 🟢 Free

The Question

An AC voltage source of RMS value $V$ and angular frequency $\omega$ is connected in series with an ideal inductor of inductance $L$ and a resistor. At the ambient temperature $T_0$ , the resistor has a resistance $R_0$ (where $R_0 < \omega L$ ). The resistance varies with temperature $T$ as $R(T) = R_0[1 + \alpha(T - T_0)]$ , where $\alpha > 0$ is the temperature coefficient of resistance. The resistor dissipates heat to the surroundings according to Newton's law of cooling with a cooling constant $K$ . If the circuit reaches a steady thermal state where the power factor of the circuit is exactly $\frac{1}{\sqrt{2}}$ , what is the required RMS voltage $V$ of the source?

Options

A.
$\sqrt{\frac{2 K \omega L (\omega L - R_0)}{\alpha R_0}}$
B.
$\sqrt{\frac{K \omega L (\omega L - R_0)}{2 \alpha R_0}}$
C.
$\sqrt{\frac{K R_0 (\omega L - R_0)}{\alpha \omega L}}$
D.
$\sqrt{\frac{2 K \omega L (\omega L + R_0)}{\alpha R_0}}$

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

Power Factor Definition|Average Power over Cycle|Internal Resistance Variation

Why This Question Is Tricky

Common Mistakes

The core misconception tested here is treating the power factor and impedance as static initial parameters rather than dynamic variables coupled to the thermodynamic state.

Students often calculate power dissipation using the cold resistance $R_0$ , failing to recognize that Joule heating alters $R(T)$ , which in turn shifts the impedance $Z(T)$ and the power factor $\cos \phi(T)$ .

This trap exposes the failure to establish a simultaneous electro-thermal equilibrium where the steady-state resistance must self-consistently satisfy both the circuit's AC power equation and Newton's law of cooling.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Power Factor Definition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free