An 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. 

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  1. 117 Ω
  2. 36 Ω 
  3. 27.2 Ω
  4. 44 Ω 

Answer (Detailed Solution Below)

Option 2 : 36 Ω 
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Detailed Solution

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The 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

X\(1 \over 2 \pi × f × C\)

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 

\(Z = \sqrt {{R^2} + {X^2}}\)

100 = 36 + X2

X = 64

X = 8

X = (XL - XC)

XL - Xc = 8

 - Xc = 8 - 43.98

Xc = 35.98 

Xc can be approximated to be 36 Ω

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