A resistor of 100 ohm ,pure inductance ofL=0.5H and capacitor are in series in a circuit containing an a.c. source of 220V and 50Hz .In the circuit current is ahead of the voltage by 30 degree . find the value of the capacitance .
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The value of the capacitance is C = 1.46 F
Explanation:
In the circuits
Inductive Reactance X2 =2πvL=2π(50)×(0.5)=50πΩ
Let capacitance be C so, xc = 1/ 2πvc = 1 / 100πc
Current leads voltage (circuit is more capacitive) by 30° = ϕ
tanϕ = X C −X L / R
tan (30°) = 1/ 100πc B − 50π / 100
That gives C = 1.46 F
Hence the value of the capacitance is C = 1.46 F
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The equivalent resistance of circuit across points A and B is equal to
[A] 22.5 Ω
[B] 25 Ω
[C] 37.5 Ω
[D] 75 Ω
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