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NEET PHYSICSALTERNATING CURRENTEasy

Question

A 40μF40 \mu\text{F} capacitor is connected to a 200 V200\text{ V}, 50 Hz50\text{ Hz} AC supply. The RMS value of the current in the circuit is, nearly:

A

2.05 A

B

2.5 A

C

25.1 A

D

1.7 A

Step-by-Step Solution

To find the RMS current (IrmsI_{rms}), we first calculate the capacitive reactance (XCX_C) using the formula XC=12πfCX_C = \frac{1}{2\pi fC}. Given C=40×10−6 FC = 40 \times 10^{-6}\text{ F} and f=50 Hzf = 50\text{ Hz}, XC=12×3.14×50×40×10−6≈79.6 ΩX_C = \frac{1}{2 \times 3.14 \times 50 \times 40 \times 10^{-6}} \approx 79.6 \ \Omega. According to the sources, the RMS current is given by Irms=VrmsXCI_{rms} = \frac{V_{rms}}{X_C} . Substituting the given RMS voltage (200 V200\text{ V}), we get Irms=20079.6≈2.51 AI_{rms} = \frac{200}{79.6} \approx 2.51\text{ A}. Therefore, the RMS current is nearly 2.5 A.

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.

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