*Why the Circuit Power factor*

*(Cos θ) Decreases, when Inductance (L) or inductive reactance (X*

_{L}) increases, In inductive circuit?

*OR**Explain the statement that “*

*the Circuit Power factor**(Cos θ) Decreases, when Inductance (L) or inductive reactance (X*_{L}) increases”

__Explanation:__

*Suppose,*

*when Inductance (L) = 0.02H*V=220, R= 10 Ω, L=0.02 H, f=50Hz.

X_{L} = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R

^{2}+X_{L}^{2}) = √ (10^{2}+ 6.28^{2}) = 11.8 ΩCos θ = R/Z = 10/11.05 = 0.85

*Now we increases Inductance (L) form 0.02 H to 0.04 H,*V=220, R= 10 Ω, L=0.04 H, f=50Hz.

X

_{L}= 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 ΩZ = √ (R

^{2}+X_{L}^{2}) = √ (10^{2}+ 12.56^{2}) = 16.05 ΩCos θ = R/Z = 10/16.05 = 0.75

*Conclusion*:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A, and Circuit power factor was (Cos θ) = 0.85.

But, when Circuit inductance increased from 0.02H to 0.04 H, then Power Factor (Cos θ) decreased from 0.85 to 0.75.

*Hence proved,*

In inductive circuit, when inductive reactance X

_{L}increases, the circuit power factor__also Decreases.__**JLCPCB - Prototype PCBs for $2 + Free Shipping on First Order**

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