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Question #42720

Difficulty: 8/10 | 🟢 Free

The Question

A resistor has a resistance $R_0 = 120 \, \Omega$ at an ambient temperature $T_0 = 300 \, \text{K}$ . Its temperature coefficient of resistance is $\alpha = 5 \times 10^{-3} \, \text{K}^{-1}$ , such that its resistance varies with temperature $T$ as $R(T) = R_0[1 + \alpha(T - T_0)]$ . The resistor is connected to a constant DC voltage source of $V = 60 \, \text{V}$ . It dissipates heat to the surroundings according to Newton's law of cooling, with a thermal conductance $K = 0.2 \, \text{W/K}$ (i.e., the rate of heat loss is $K(T - T_0)$ ). What is the steady-state electrical power dissipated by the resistor?

Options

A.
$20 \, \text{W}$
B.
$25 \, \text{W}$
C.
$30 \, \text{W}$
D.
$36 \, \text{W}$

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

Ohm's Law Validity|Power Dissipation Ratio|Thermal Conductivity in Series

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize and model thermal feedback in electrical circuits, a classic non-ohmic steady-state problem.

The primary trap is treating resistance as a constant scalar ( $R = R_0$ ) and directly applying $P = V^2/R_0$ , which incorrectly yields $30 \, \text{W}$ .

Students fall into this trap by ignoring the fact that power dissipation increases the resistor's temperature, which in turn increases its resistance and reduces the power drawn from a constant voltage source.

Solving it requires coupling Ohm's Law with Newton's Law of Cooling to find the true equilibrium state.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Ohm's Law Validity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free