a load consisting of a capacitor in series with a resistor has an impedance of 50Ω and pf of 0.707 leading. The lead is connected in series with a 40Ω resistor across ac supply and the resulting current is 3A. Determine the supply voltage and overall phase angle.
Answers
Answer:
The supply voltage and overall phase angle is 249.69 V, 25.135.
Explanation:
A resistor and a capacitor are connected in series to a 50 Hz ac source.
The rms voltage across the resistor = = 151 volt
The rms voltage across the capacitor = = 160.3 volt
Let The RMS voltage of the source = V volt
Or, V = √48497.09
V = 220.22
So, The RMS voltage of the source = V = 220.22 volt
i.e, Impedance of circuit = X = 291.68
So, The capacitive reactance = 212.3 ohm
Hence, The RMS voltage of the source is 220.22 volt
Impedance of circuit is 291.68 ohm.
The capacitive reactance is 212.3 ohm.
Given,
Resistance, R=30Ω
Inductive Reactance, X
=20Ω
Frequency, f=50Hz
We know,
= ωL = 2πfL
20 = X
= 2π.50.L
When frequency of ac source is changed to 100Hz,
L=2π.100.L
=2π.(50×2)L
=20×2Ω (from )
L = 40Ω
Impedance, Z = (40) 2+(30) 2
Z=50Ω
Current, I = Z
V = 50
200 = 4A
Hence, the supply voltage and overall phase angle is 249.69 V, 25.135.
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