# In an Inductive Circuit, Why the Current Increases When Frequency Decreases?

**Why the Current (I) Decreases, When Frequency Increases in an Inductive Circuit & vise versa?**

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

Explain the statement that “**In an Inductive circuit, why the circuit current increases when frequency decreases**“.

*Related Questions:*

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

**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 inductance “L” as well as inductive reactance “X_{L}” as inductance and inductive reactances are directly proportional to each others.

Let’s check with an example to see how current reduced by increase in frequency in case of inductive circuit.

__When Frequency = 50 Hz__

Suppose an inductive circuit where:

- Voltage = V = 3000 V
- Inductance = L = 0.1 Henry
- Resistance = R = 12 Ω
- Frequency =
*f*= 50 Hz

To find the inductive reactance;

X_{L} = 2π*f*L

X_{L} = 2 x 3.1415 x 50 x 0.1

X_{L} = 31.415 Ω

Now circuit impedance:

Z = √ (R^{2 }+ X_{L}^{2})

Z = √ (12^{2} + 31.415^{2})

Z = 33.63 Ω

Finally, current in inductive circuit:

I = V / Z

I = 3000 V / 33.63 Ω

** I = 89.20 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 Frequency = 60 Hz__

*Now we increased the frequency form 50Hz to 60Hz.*

V = 3kV, R = 12 Ω, L = 0.1 H, *f* = 60 Hz.

X_{L} = 2π*f*L= 2 x 3.1415 x 60 x 0.1 = 37.7 Ω

Z = √ (R^{2 }+ X_{L}^{2}) = √ (12^{2} + 37.7^{2}) = 39.56 Ω

I = V / Z = 3 kV / 39.56 Ω

** I = 75.83 A **

**Conclusion:**

We can see that, When frequency was **50Hz**, then the circuit current were **89.20 A**,

But when circuit frequency increased from **50Hz** to **60Hz**, then the current decreased from **89.20 A** to **75.83 A**.

**Hence proved,**

**In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.**

*f *∝ 1 / I

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 and frequency.

**L ****∝**** f** and **L ****∝**** X _{L}**

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

**I ****∝**** 1 / L and I ****∝**** 1 / X _{L} and I **

**∝**

**1 / Z**

- Impedance is directly proportional to the inductive reactance

**Z ****∝**** X _{L}**

**In an inductive circuit, Frequency is inversely proportional to the current**

**I ****∝**** 1 /*** f*

Related Questions/Answers:

- Why Flux in Primary and Secondary Winding is Always Equal?
- Why Power in Pure Inductive and Pure Capacitive Circuit is Zero?
- 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)?

In inductive ckt

I=v/z

I= 3000/33.63

=89.26 A not 82.20

Thanks for correction. It was type and has been corrected now.

What are the roles of capacitors in electric or electronic circuits ?