# Why Current Decreases When Inductance or Inductive Reactance Increases?

**In Inductive Circuit, Why the Circuit Current (I) Decreases, When Inductance (L) or Inductive reactance (X**_{L}) Increases?

_{L}) Increases?

Another question from electrical and electronics engineering interviews question and answers series.

Explain the statement that ” **In an Inductive circuit, when Inductance (L) or inductive reactance (X _{L}) increases, the Circuit Current (I) decreases**“.

*Related Question: Why Current Increases When Capacitance Increases or Capacitive Reactance Decreases?*

**Explanation:**

We know that in DC circuits:

I = V / R,

But in case of AC circuits:

I = V / Z

Where “total resistance of AC circuits = Impedance = Z = √ (R^{2} + (X_{L} – X_{C}^{2})”

In case of Inductive circuit:

- Z = √ (R
^{2}+ X_{L}^{2}) - I = V / X
_{L}or I = V / Z

It shows that in inductive circuit, Current is inversely proportional to the inductive reactance as well as inductance “L” as inductance and inductive reactances “X_{L}” are directly proportional to each others.

Let’s check with an example to see how current reduced by inductive reactance.

**When Inductance = 0.02 H**

Suppose an inductive circuit where:

- Inductance = L = 0.02 Henry
- Voltage = V = 220 V
- Resistance = R = 10 Ω
- Frequency =
*f*= 50 Hz

To find the inductive reactance;

X_{L} = 2π*f*L

X_{L} = 2 x 3.1415 x 50 x 0.02

X_{L} = 6.28 Ω

Now circuit impedance:

Z = √ (R^{2 }+ X_{L}^{2})

Z = √ (10^{2} + 6.28^{2})

Z = 11.8 Ω

Finally, current in inductive circuit:

I = V / Z

I = 220 V / 11.8 Ω

I = 18.64 A

Related Questions:

- Why Power Factor Decreases When Inductance or Inductive Reactance Increases?
- Why Power Factor Decreases When Capacitive Reactance Increases or Capacitance Decreases?

**When Inductance = 0.04 H**

Now we increased the Inductance (L) of inductor form 0.02 H to 0.04 H,

V = 220 V, R = 10 Ω, L = 0.04 H, *f* = 50 Hz.

X_{L} = 2π*f*L= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R^{2 }+ X_{L}^{2}) = √ (10^{2} + 12.56^{2}) = 16.05 Ω

I = V / Z = 220 V / 16.05 Ω

I = 13.70 A

Related Questions:

- In a Capacitive Circuit, Why the Current Increases When Frequency Increases?
- In an Inductive Circuit, Why the Current Increases When Frequency Decreases?

**Conclusion:**

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

But when circuit inductance increased from **0.02 H** to **0.04 H**, then the current decreased from **18.64 A** to **13.70 A**.

**Hence proved,**

**In an inductive circuit, when inductive reactance X _{L} increases, the circuit current decreases and vice versa.**

In oral or verbal,

- Inductive reactance is a kind of resistance. When resistance increases, the circuit current decreases and vice versa.
- Inductance is directly proportional to the inductive reactance

**L ∝ X _{L}**

- Current is inversely proportional to the inductance and inductive reactance.

**I ∝ 1/L and I ∝ 1/X _{L}**

Related Questions/Answers:

- Why Flux in Primary and Secondary Winding is Always Equal?
- Why the reactance of a system under fault condition is low and faults currents may raise to the dangerously high value?
- Why Inductive Reactance (X
_{L}In DC Supply Is Zero (0)?

Thanks for easy explanation.

Hello,

thank u for valuable information.

my query is,

why we don’t use below 50 Hz in e-machines?

Plz give answer.

Thanks sir for providing this type knowledgeable questions.

Pls continue it sir ………….