Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $I$ and voltage across the capacitor is $V$ , then:

V_a > V_b

V_a < V_b

V_a = V_b

V_a = V_b = 0

Step-by-Step Solution

Capacitance: When a dielectric like mica (with dielectric constant $K > 1$ ) fills the capacitor, the capacitance increases from $C_a$ to $C_b = K C_a$ .
Reactance: Capacitive reactance is inversely proportional to capacitance ( $X_C = 1/\omega C$ ). Since $C$ increases, $X_C$ decreases ( $X_{C,b} < X_{C,a}$ ) .
Impedance and Current: The total impedance $Z = \sqrt{R^2 + X_C^2}$ decreases. Consequently, the current in the circuit $I = V_{source}/Z$ increases ( $I_b > I_a$ ).
Voltage across Capacitor: The voltage across the capacitor is given by $V_C = I X_C = \frac{V_{source} X_C}{\sqrt{R^2 + X_C^2}} = \frac{V_{source}}{\sqrt{(R/X_C)^2 + 1}}$ . As $C$ increases, $X_C$ decreases, causing the term $(R/X_C)$ to increase. This increases the denominator, thereby decreasing the voltage across the capacitor. Thus, $V_a > V_b$ .

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