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Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL) increases in inductive circuit?

 Explain the statement that ” In Inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit Current (I) decrease”
OR
Why the circuit Current (I) decrease, when  Inductance (L) or inductive reactance (XL) increases in inductive circuit?

Explanation:
We know that, I = V / R, 
but in inductive circuit, I = V/XL
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.

XL = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R2+XL2) = √ (102 + 6.282) = 11.8 Ω
I = 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.
XL = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω
Z = √ (R2+XL2) = √ (102 + 12.562) = 16.05 Ω
I = 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 XL increases, the circuit current decreases, and Vice Virsa.

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5 Comments

    1. Electrical Technology says:

      Welcome Dear

  1. Aamir agwan says:

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    1. Electrical Technology says:

      Thanks Dear

  2. PATIL VAIBHAV BHASKAR says:

    Hello,
    thank u for valuable informatin.
    my query is,
    why we dnt use below 50 Hz in e-machines?
    Pls give answer.

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