Question
Download Solution PDFAn inductor having an inductance of 0.14 H and a resistance of 6 Ω is connected in series with a capacitor across 50 Hz Supply. Calculate the capacitive reactance required to give the circuit power factor 0.6 lagging. Assume supply voltage to be sinusoidal.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 2):(36 Ω )
Concept:
The net impedance for a series circuit is given by:
Z = R + j (XL - XC)
\(Z = \sqrt {{R^2} + {X^2}}\)
R = Net resistance of the series circuit.
XL = Net inductive Reactance defined as: XL = ωL =
XC = Net capacitive Reactance defined as
XC = \(1 \over 2 \pi × f × C\)
Power factor \(= \frac{R}{Z}\)
f is the frequency in Hz
C is the capacitance in F
L is the inductance in H
Calculation:
Given
L = 0.14 H
R = 6 Ω
Power factor = 0.6
\(Z = \sqrt {{R^2} + {X^2}}\)
XL = 2 × 50 × 0.14 × 3.14
= 43.98
Power factor \(= \frac{R}{Z}\)
0.6 = \(6 \over Z\)
Z = 10 Ω
\(Z = \sqrt {{R^2} + {X^2}}\)
100 = 36 + X2
X2 = 64
X = 8
X = (XL - XC)
XL - Xc = 8
- Xc = 8 - 43.98
Xc = 35.98 Ω
Xc can be approximated to be 36 Ω
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