Q20. A 20-watt, 50 V lamp is to be connected to AC mains of 250 V, 50 Hz. Calculate the value of ser

Q: What is the correct answer for CUET UG Physics Q20?

The correct answer is d. First, determine the operating current and resistance of the lamp:Power (P) = 20 W, Rated Voltage (V_lamp) = 50 V.Current (I) = P / V_lamp = 20 W / 50 V = 0.4 A. This is the current required for the lamp to operate normally.Resistance of the lamp (R) = V_lamp / I = 50 V / 0.4 A = 125 Ω.When connected to AC mains, the current through the series circuit (lamp and capacitor) must be 0.4 A for the lam

Q: Q20. A 20-watt, 50 V lamp is to be connected to AC mains of 250 V, 50 Hz. Calculate the value of series capacitor to run the lamp. (a) 1.4 µF (b) 2.1 µF (c) 3.5 µF (d) 4.9 µF

The correct answer is d. First, determine the operating current and resistance of the lamp:Power (P) = 20 W, Rated Voltage (V_lamp) = 50 V.Current (I) = P / V_lamp = 20 W / 50 V = 0.4 A. This is the current required for the lamp to operate normally.Resistance of the lamp (R) = V_lamp / I = 50 V / 0.4 A = 125 Ω.When connected to AC mains, the current through the series circuit (lamp and capacitor) must be 0.4 A for the lam

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

A. 1.4 µF · B. 2.1 µF · C. 3.5 µF · D. 4.9 µF

Question

Q20. A 20-watt, 50 V lamp is to be connected to AC mains of 250 V, 50 Hz. Calculate the value of series capacitor to run the lamp.
(a) 1.4 µF
(b) 2.1 µF
(c) 3.5 µF
(d) 4.9 µF

FirstInTest · Accepted Answer

First, determine the operating current and resistance of the lamp:Power (P) = 20 W, Rated Voltage (V_lamp) = 50 V.Current (I) = P / V_lamp = 20 W / 50 V = 0.4 A. This is the current required for the lamp to operate normally.Resistance of the lamp (R) = V_lamp / I = 50 V / 0.4 A = 125 Ω.When connected to AC mains, the current through the series circuit (lamp and capacitor) must be 0.4 A for the lamp to operate as intended. The AC mains voltage is V_rms = 250 V, and frequency (f) = 50 Hz.The total impedance (Z) of the series RC circuit (lamp resistance and capacitor) is Z = V_rms / I = 250 V / 0.4 A = 625 Ω.In a series RC circuit, the impedance is given by Z = √(R² + X_C²), where X_C is the capacitive reactance.So, X_C = √(Z² - R²) = √(625² - 125²) = √(390625 - 15625) = √375000 Ω.√375000 ≈ 612.37 Ω.Now, relate capacitive reactance to capacitance: X_C = 1 / (2πfC).So, C = 1 / (2πfX_C) = 1 / (2 * 3.14159 * 50 Hz * 612.37 Ω) = 1 / (192301.6) F ≈ 5.20 × 10⁻⁶ F = 5.20 µF.The closest option is (d) 4.9 µF.

Answer

1.4 µF

Answer

2.1 µF

Answer

3.5 µF

Answer

4.9 µF

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