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NEET PHYSICSALTERNATING CURRENTMedium

Question

A 100 Ω100 \, \Omega resistance and a capacitor of 100 Ω100 \, \Omega reactance are connected in series across a 220220 V source. When the capacitor is 50%50\% charged, the peak value of the displacement current is:

A

2.2 A

B

11 A

C

4.4 A

D

11\sqrt{2} A

Step-by-Step Solution

The displacement current (IdI_d) in the capacitor is equal to the conduction current (IcI_c) flowing in the series circuit. In an AC circuit, the peak displacement current corresponds to the peak conduction current (I0I_0). Given: Resistance R=100 ΩR = 100 \, \Omega, Capacitive Reactance XC=100 ΩX_C = 100 \, \Omega, Source Voltage Vrms=220V_{rms} = 220 V. The impedance of the series RC circuit is Z=R2+XC2=1002+1002=1002 ΩZ = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \, \Omega . The peak voltage of the source is V0=2Vrms=2202V_0 = \sqrt{2} V_{rms} = 220\sqrt{2} V. The peak current is I0=V0Z=22021002=2.2I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2 A . The condition '50% charged' refers to the instantaneous state, but the peak value of the current is a constant property of the steady-state AC circuit.

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 CURRENTresistancecapacitorreactanceconnectedseries

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