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NEET PHYSICSALTERNATING CURRENTMedium

Question

A 10 μ\muF capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (π=3.14\pi = 3.14):

A

0.93 A

B

1.20 A

C

0.35 A

D

0.58 A

Step-by-Step Solution

To find the peak current (imi_m), we first calculate the peak voltage (VmV_m) and the capacitive reactance (XCX_C).

  1. Peak Voltage (VmV_m): The given source voltage is the RMS value (Vrms=210V_{rms} = 210 V). According to the sources, Vm=Vrms2=210×1.414≈296.94V_m = V_{rms}\sqrt{2} = 210 \times 1.414 \approx 296.94 V .
  2. Capacitive Reactance (XCX_C): XC=1ωC=12πfCX_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}. Substituting the values: XC=12×3.14×50×10×10−6=1314×10−5≈318.47 ΩX_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1}{314 \times 10^{-5}} \approx 318.47 \ \Omega .
  3. Peak Current (imi_m): im=VmXC=VmωC=296.94×(314×10−5)≈0.932i_m = \frac{V_m}{X_C} = V_m \omega C = 296.94 \times (314 \times 10^{-5}) \approx 0.932 A. Thus, the peak current is nearly 0.93 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.

PHYSICSALTERNATING CURRENTcapacitorconnectedsourcecurrentcircuit

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