Q47. A bulb and a capacitor are connected in series to an a.c. source. A dielectric slab is now intr

Q: What is the correct answer for CUET UG Physics Q47?

The correct answer is B. When a dielectric slab is introduced between the plates of a capacitor, its capacitance (C) increases by a factor equal to the dielectric constant (k) of the material (C' = kC, where k > 1). In an AC circuit, a capacitor offers capacitive reactance (X_C) which is given by X_C = 1/(ωC), where ω is the angular frequency of the AC source. If the capacitance C increases, the capacitive reactance X_C d

Q: Q47. A bulb and a capacitor are connected in series to an a.c. source. A dielectric slab is now introduced between the plates of the capacitor. The intensity of the bulb will be: (a) Zero (b)…

The correct answer is B. When a dielectric slab is introduced between the plates of a capacitor, its capacitance (C) increases by a factor equal to the dielectric constant (k) of the material (C' = kC, where k > 1). In an AC circuit, a capacitor offers capacitive reactance (X_C) which is given by X_C = 1/(ωC), where ω is the angular frequency of the AC source. If the capacitance C increases, the capacitive reactance X_C d

Q: What are the options for this CUET UG Physics question?

A. Zero · B. Increases · C. Decreases · D. Remain same

Question

Q47. A bulb and a capacitor are connected in series to an a.c. source. A dielectric slab is now introduced between the plates of the capacitor. The intensity of the bulb will be:
(a) Zero
(b) Increases
(c) Decreases
(d) Remain same

FirstInTest · Accepted Answer

When a dielectric slab is introduced between the plates of a capacitor, its capacitance (C) increases by a factor equal to the dielectric constant (k) of the material (C' = kC, where k > 1). In an AC circuit, a capacitor offers capacitive reactance (X_C) which is given by X_C = 1/(ωC), where ω is the angular frequency of the AC source. If the capacitance C increases, the capacitive reactance X_C decreases. The total impedance (Z) of the series R-C circuit, where the bulb has resistance R, is given by Z = √(R² + X_C²). Since X_C decreases, the total impedance Z of the circuit decreases. The current (I) flowing through the circuit is I = V_rms / Z, where V_rms is the RMS voltage of the AC source. As Z decreases, the current I increases. The brightness (intensity) of the bulb is proportional to the power dissipated by it, which is given by I²R. Since the current I increases, the power dissipated by the bulb increases, making it glow brighter. Therefore, the intensity of the bulb increases.

Answer

Zero

Answer

Increases

Answer

Decreases

Answer

Remain same

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