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“.In Inductive Circuit, Why Current Increases When Frequency Decreases

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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 = √ (R2 + (XL – XC2)”

In case of Inductive circuit:

It shows that in inductive circuit, Current is inversely proportional to the inductance “L” as well as inductive reactance “XL” 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:

To find the inductive reactance;

XL = 2πfL

XL = 2 x 3.1415 x 50 x 0.1

XL = 31.415 Ω

Now circuit impedance:

Z = √ (R2 + XL2)

Z = √ (122 + 31.4152)

Z = 33.63 Ω

Finally, current in inductive circuit:

I = V / Z

I = 3000 V / 33.63 Ω

 I = 89.20 A 

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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.

XL = 2πfL= 2 x 3.1415 x 60 x 0.1 = 37.7 Ω

Z = √ (R2 + XL2) = √ (122 + 37.72) = 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,

L f    and    L XL

I 1 / L     and     I 1 / XL    and I 1 / Z

Z XL

I 1 / f

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