Before You Begin

Question #43450

Difficulty: 8/10 | 🟢 Free

The Question

A full-wave rectifier circuit operates with an input AC frequency $f = 50 \text{ Hz}$ and produces a peak output voltage $V_m = 20 \text{ V}$ . The output is filtered by a capacitor $C = \frac{250}{\sqrt{3}} \mu\text{F}$ and connected to a load which is a Positive Temperature Coefficient (PTC) thermistor. The thermistor's resistance varies with its temperature $T$ as $R(T) = R_0 [1 + \alpha (T - T_a)]$ , where $R_0 = 100 \, \Omega$ is the resistance at the ambient temperature $T_a$ , and $\alpha = 0.05 \text{ K}^{-1}$ . The thermistor dissipates heat to the surroundings at a rate $P_{loss} = K(T - T_a)$ , where $K = 0.1 \text{ W/K}$ . Assuming the DC output voltage across the load is approximately equal to $V_m$ for power calculations, what is the steady-state ripple factor $\gamma$ of the output voltage?

Options

A.
$\frac{1}{10}$
B.
$\frac{1}{5}$
C.
$\frac{1}{15}$
D.
$\frac{1}{20}$

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

Rectifier Ripple Factor|Temperature Coefficient (Alpha)|Power Dissipation Ratio

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle coupled electro-thermal systems, a common hallmark of high-tier JEE Advanced multi-concept problems.

The primary trap is treating the load resistance $R$ as a constant $R_0$ (which incorrectly yields $\gamma = 1/5$ ), failing to recognize that the power dissipated by the rectifier alters the thermistor's temperature, which in turn alters the resistance and the filtering requirements.

A secondary trap involves calculating the power dissipation using the initial resistance ( $P = V_m^2/R_0$ ) and assuming it stays constant to find $\Delta T$ , which ignores the dynamic feedback loop and leads to an incorrect steady-state resistance of $300 \, \Omega$ (yielding $\gamma = 1/15$ ).

Students must correctly apply steady-state thermal equilibrium to establish and solve the governing quadratic equation for $R$ before evaluating the ripple factor.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Semiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
Concept	Rectifier Ripple Factor
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free