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NEET PHYSICSALTERNATING CURRENTEasy

Question

A coil of self-inductance L is connected in series with a bulb B and an AC source. The brightness of the bulb decreases when:

A

Frequency of the AC source is decreased

B

The number of turns in the coil is reduced

C

A capacitance of reactance X_C = X_L is included in the same circuit

D

An iron rod is inserted in the coil

Step-by-Step Solution

The brightness of the bulb depends on the RMS current flowing through it (P=I2RP = I^2R). The current in the series circuit is given by I=V/ZI = V/Z, where Z=R2+XL2=R2+(2πfL)2Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (2\pi f L)^2}. To decrease the brightness, the current must decrease, which implies the impedance ZZ (and thus the inductive reactance XLX_L) must increase.

  • Option 1: Decreasing frequency ff decreases XLX_L, which increases II and brightness.
  • Option 2: Reducing the number of turns NN decreases self-inductance LL (since L∝N2L \propto N^2 for a solenoid), which decreases XLX_L and increases brightness .
  • Option 3: Adding a capacitor where XC=XLX_C = X_L brings the circuit to resonance, minimizing impedance (Z=RZ=R) and maximizing current and brightness.
  • Option 4: Inserting an iron rod increases the relative permeability μr\mu_r of the core material. Since L=μrμ0n2AlL = \mu_r \mu_0 n^2 Al, the inductance LL increases significantly . This increases XLX_L and ZZ, thereby decreasing the current and the bulb's brightness.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTselfinductanceconnectedseriessourcebrightness

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