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NEET PHYSICSALTERNATING CURRENTMedium

Question

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series to an AC source of potential difference V volts. The potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of the current flowing through the LCR series circuit is 10210\sqrt{2} A. The impedance of the circuit will be:

A

4 Ω\Omega

B

5 Ω\Omega

C

42 Ω4\sqrt{2} \ \Omega

D

52 Ω5\sqrt{2} \ \Omega

Step-by-Step Solution

To find the impedance (ZZ), we first determine the total RMS voltage (VV) of the source. In a series LCR circuit, this is given by the phasor sum: V=VR2+(VL−VC)2V = \sqrt{V_R^2 + (V_L - V_C)^2}. Substituting the given values, V=402+(40−10)2=1600+900=50V = \sqrt{40^2 + (40 - 10)^2} = \sqrt{1600 + 900} = 50 V . The RMS current (II) is calculated from the peak current amplitude (imi_m) using the relationship I=im2I = \frac{i_m}{\sqrt{2}}. Given im=102i_m = 10\sqrt{2} A, I=1022=10I = \frac{10\sqrt{2}}{\sqrt{2}} = 10 A . The impedance is then the ratio of the RMS voltage to the RMS current: Z=VI=50 V10 A=5 ΩZ = \frac{V}{I} = \frac{50 \text{ V}}{10 \text{ A}} = 5 \ \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 CURRENTinductorinductancecapacitorcapacitanceresistor

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