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NEET PHYSICSALTERNATING CURRENTMedium

Question

In the given circuit, the reading of voltmeter V1 and V2 are 300 V each. The reading of the voltmeter V3 and ammeter A are respectively:

A

150 V, 2.2 A

B

220 V, 2.2 A

C

220 V, 2.0 A

D

100 V, 2.0 A

Step-by-Step Solution

In a series LCR circuit, the total source voltage VV is related to the voltages across the resistor (VRV_R, read by V3V_3), inductor (VLV_L, read by V1V_1), and capacitor (VCV_C, read by V2V_2) by the phasor relationship: V=VR2+(VLVC)2V = \sqrt{V_R^2 + (V_L - V_C)^2} . Given that V1=V2=300V_1 = V_2 = 300 V, the circuit is in a state of resonance where the inductive and capacitive voltages cancel each other out (VLVC=0V_L - V_C = 0). Therefore, V=VR2=VRV = \sqrt{V_R^2} = V_R. This implies the reading of voltmeter V3V_3 is equal to the source voltage. Based on the options and standard mains voltage conventions (and the specific data of this AIPMT 2010 problem where source is 220V, 50Hz, R=100\Omega ), the source voltage is 220 V. Thus, V3=220V_3 = 220 V. The current in the circuit at resonance is determined by the resistance: I=V/RI = V/R. Using the probable answer values (I=2.2I = 2.2 A) and derived voltage (V=220V = 220 V), this is consistent with a resistance R=100ΩR = 100 \, \Omega.

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 CURRENTcircuitreadingvoltmeterreadingvoltmeter

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