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NEET PHYSICSALTERNATING CURRENTEasy

Question

A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is:

A

series LR

B

series RC

C

series LC

D

series LCR

Step-by-Step Solution

  1. DC Behavior: When connected to a DC source, the frequency ω\omega is zero. A capacitor offers infinite reactance (XC=1/ωC=∞X_C = 1/\omega C = \infty) to DC, blocking the current completely . Since a steady current of 0.4 A flows, the circuit cannot contain a capacitor in series. This eliminates options containing C (RC, LC, LCR).
  2. Resistance: The resistance of the circuit can be calculated using the DC data: R=VDC/IDC=12 V/0.4 A=30 ΩR = V_{DC} / I_{DC} = 12 \text{ V} / 0.4 \text{ A} = 30 \, \Omega .
  3. AC Behavior: When connected to an AC source, the circuit offers impedance ZZ. Calculating impedance: Z=VAC/IAC=12 V/0.2 A=60 ΩZ = V_{AC} / I_{AC} = 12 \text{ V} / 0.2 \text{ A} = 60 \, \Omega.
  4. Conclusion: The impedance (60 Ω60 \, \Omega) is greater than the resistance (30 Ω30 \, \Omega). This indicates the presence of a reactive component (XX) such that Z=R2+X2Z = \sqrt{R^2 + X^2}. Since capacitors are ruled out, the reactive component must be an inductor (XLX_L). Thus, the circuit is a series combination of Inductance (L) and Resistance (R) .

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 CURRENTcircuitconnectedsourcecurrentcircuit

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