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NEET PHYSICSALTERNATING CURRENTMedium

Question

A series R-C circuit is connected to an alternating voltage source. Consider two situations: 1. When the capacitor is air-filled. 2. When the capacitor is mica filled. If the current through the resistor is II and voltage across the capacitor is VV, then:

A

V_a > V_b

B

V_a < V_b

C

V_a = V_b

D

V_a = V_b = 0

Step-by-Step Solution

  1. Capacitance: When a dielectric like mica (with dielectric constant K>1K > 1) fills the capacitor, the capacitance increases from CaC_a to Cb=KCaC_b = K C_a .
  2. Reactance: Capacitive reactance is inversely proportional to capacitance (XC=1/ωCX_C = 1/\omega C). Since CC increases, XCX_C decreases (XC,b<XC,aX_{C,b} < X_{C,a}) .
  3. Impedance and Current: The total impedance Z=R2+XC2Z = \sqrt{R^2 + X_C^2} decreases. Consequently, the current in the circuit I=Vsource/ZI = V_{source}/Z increases (Ib>IaI_b > I_a).
  4. Voltage across Capacitor: The voltage across the capacitor is given by VC=IXC=VsourceXCR2+XC2=Vsource(R/XC)2+1V_C = I X_C = \frac{V_{source} X_C}{\sqrt{R^2 + X_C^2}} = \frac{V_{source}}{\sqrt{(R/X_C)^2 + 1}}. As CC increases, XCX_C decreases, causing the term (R/XC)(R/X_C) to increase. This increases the denominator, thereby decreasing the voltage across the capacitor. Thus, Va>VbV_a > V_b.

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 CURRENTseriescircuitconnectedalternatingvoltage

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