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NEET PHYSICSALTERNATING CURRENTEasy

Question

In an AC circuit, an alternating voltage ε=2002sin(100t)\varepsilon = 200\sqrt{2}\sin(100t) V is connected to a capacitor of capacity 1μF1 \mu\text{F}. The RMS value of the current in the circuit is:

A

100 mA

B

200 mA

C

20 mA

D

10 mA

Step-by-Step Solution

  1. Identify Parameters: Compare the given voltage equation ε=2002sin(100t)\varepsilon = 200\sqrt{2}\sin(100t) with the standard form V=Vmsin(ωt)V = V_m \sin(\omega t). Peak Voltage Vm=2002V_m = 200\sqrt{2} V. Angular Frequency ω=100\omega = 100 rad/s.
  • Capacitance C=1μF=1×106C = 1 \mu\text{F} = 1 \times 10^{-6} F.
  1. Calculate Reactance: The capacitive reactance is XC=1ωC=1100×106=1104=104ΩX_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \, \Omega .
  2. Calculate RMS Voltage: Vrms=Vm2=20022=200V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 V .
  3. Calculate RMS Current: Irms=VrmsXC=200104=0.02I_{rms} = \frac{V_{rms}}{X_C} = \frac{200}{10^4} = 0.02 A.
  4. Convert Units: 0.02 A=200.02 \text{ A} = 20 mA.

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 CURRENTcircuitalternatingvoltagevarepsilonsqrtsint

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