*Explain the statement that ” In Inductive circuit, when Inductance (L) or inductive reactance (X*

_{L}) increases, the circuit Current (I) decrease”**OR**

*Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (X*

_{L}) increases in inductive circuit?__Explanation:__

We know that, I = V / R,

but in inductive circuit, I = V/X_{L}_{So Current in inversely proportional to the Current ( in inductive circuit.}_{Let‘s check with an example.. }*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 ΩI = V/Z = 220/11.8 = 18.64 A

*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 ΩI = V/Z = 220 / 16.05 = 13.70 A

**Conclusion**:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70 A to 18.64A.

*Hence proved,*In inductive circuit, when inductive reactance X

_{L}increases, the circuit current decreases, and Vice Virsa.**From Schematic to PCB Production, Just Need One Tool - EasyEDA**

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Thanks Dear

Hello,

thank u for valuable informatin.

my query is,

why we dnt use below 50 Hz in e-machines?

Pls give answer.