An alternating source of emf 12 V and frequency 50 Hz is applied to a capacitor of capacitance 3 μF in series with a resistor of resistance 1 kΩ , Calculate the phase angle. (Ans: 46.7°)



Data Given:


Voltage = V = 12 V

Frequency = f = 50 Hz

Capacitance of capacitor = C = 3 μF = 3 ｘ10⁻⁶ F

Resistance = R =  1 kΩ = 1 ｘ10³ Ω 




To Find:


Phase angle  = θ = ?


Solution:


The phase angle θ  can be calculated as

θ = tan⁻¹ (`\frac {1}{ധCR}`)

or

θ = tan⁻¹ (`\frac {1}{2ㄫfCR}`)

By putting values

θ = tan⁻¹ (`\frac {1}{2 ｘ 3.1415 ｘ50 Hz ｘ 3 ｘ10⁻⁶ F ｘ 1 ｘ10³ Ω}`)

θ = tan⁻¹ (`\frac {1}{942.48 ｘ10⁻³}`)

θ = tan⁻¹ (0.0010610 ｘ10³)

or

θ = tan⁻¹ (1.0610 )

θ = 46.7° -----------------Ans.




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