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

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Question added by muhammad arif , Assistant Engineer , Phoenix armour
Date Posted: 2013/11/23
muhammad arif
by muhammad arif , Assistant Engineer , Phoenix armour

We know that, I = V / R, but in inductive circuit, I = V/XLSo 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 x3.1415 x50 x0.02 =6.28 Ω 

Z = √ (R2+XL2) = √ (102 +6.282) =11.8 Ω I = V/Z =220/11.8 = 18.64 A

 

Vaishnavi Anil
by Vaishnavi Anil , Product Planner , Tanushh Electronics

V=Ldi/dt

where,

V=voltage across inductor

L=inductance

i=current across inductor

From the above relation ,it is understood that on integrating this equation, we obtain that current i is inversely proportional to the inductance L. Thus, when inductance increases, current increases.

Now, inductive reactance XL and inductance L are related by the equation:

XL=2πfL

which means that the two are directly related , so current also increases if XL increases.

The relation between current,voltage across the inductor and inductance is given by the formula: V=Ldi/dt V/L=di/dt. -(1) From eq1 we get L is inversely proportional to current for a constant voltage therefore when L increases current decreases.