Start Learning Free

Before You Begin

Question #43037

Difficulty: 8/10 | 🟢 Free

The Question

An AC generator consists of a flat coil of NN turns and area AA, rotating at a constant angular velocity ω\omega in a uniform magnetic field of induction BB. The axis of rotation is perpendicular to the magnetic field. The internal resistance and self-inductance of the generator coil are negligible. The output of the generator is connected across a galvanometer having a resistance RR and self-inductance LL. To protect the galvanometer, a pure resistor of resistance RSR_S is connected in parallel with it as a shunt. The value of RSR_S is chosen such that the average electrical power dissipated in the shunt is exactly equal to the average electrical power dissipated in the galvanometer. Determine the average mechanical torque τ\tau that an external agent must apply to the generator coil to maintain its constant angular velocity ω\omega. (Assume no mechanical friction).

Options

  • A.

    N2A2B2ωRR2+ω2L2\frac{N^2 A^2 B^2 \omega R}{R^2 + \omega^2 L^2}

  • B.

    N2A2B2ω(2R2+ω2L2)2R(R2+ω2L2)\frac{N^2 A^2 B^2 \omega (2R^2 + \omega^2 L^2)}{2R(R^2 + \omega^2 L^2)}

  • C.

    N2A2B2ωR\frac{N^2 A^2 B^2 \omega}{R}

  • D.

    N2A2B2ωR2(R2+ω2L2)\frac{N^2 A^2 B^2 \omega R}{2(R^2 + \omega^2 L^2)}

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

Shunt Resistance Conversion|AC Generator Back-EMF|Power Dissipation Ratio

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from DC circuit logic to AC impedance matching while applying the principle of energy conservation (Lenz's Law).

The primary trap lies in the calculation of the shunt resistance RSR_S.

In a DC circuit, equal power dissipation in parallel branches implies equal resistances (RS=RR_S = R).

However, in an AC circuit, the galvanometer's self-inductance creates a reactive back-EMF, increasing its impedance to R2+ω2L2\sqrt{R^2 + \omega^2 L^2}.

Students who incorrectly apply DC logic will assume RS=RR_S = R, leading to an incorrect total equivalent admittance and a wrong mechanical torque (Option B).

Furthermore, the problem elegantly couples AC circuit theory with rotational mechanics, requiring students to equate the average mechanical power (τω\tau \omega) to the total active electrical power dissipated, reinforcing that reactive power does not contribute to the average mechanical work required to rotate the generator.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptShunt Resistance Conversion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree