In a Capacitive Circuit, Why the Current Increases When Frequency Increases?

Why the Current (I) Increases, When Frequency Increases in a Capacitive Circuit & Vice Versa?

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

Explain the statement that “In a capacitive circuit, why the circuit current increases when frequency increases“.In a Capacitive Circuit, Why the Current Increases When Frequency Increases

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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 a capacitive circuit:

It shows that in a capacitive circuit, Current is directly proportional to the capacitance “C” and inversely proportional to the capacitive reactance as capacitance and capacitive reactances “XC” are inversely proportional to each others.

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Let’s check with an example to see how current increases by increase in frequency in case of a capacitive circuit.

When Frequency = 5 µF

Suppose a capacitive circuit where:

To find the capacitive reactance;

XC = 1 / 2πfC

XC = 1 / (2 x 3.1415 x 50 x 5×10-6)

XC = 636.94 Ω

Now, current in the capacitive circuit:

I = V / XC

I = 3000 V / 636.94 Ω

 I = 4.71 A 

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When Frequency = 60 Hz

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

V = 3kV,  C = 5 µ Farads,  f = 60 Hz.

XC = 1 / 2πfC = 1 / (2 x 3.1415 x 60 x 5×10-6) = 530.63Ω

I = V / XC = 3 kV / 530.63Ω

 I = 5.65 A 

Conclusion:

We can see that, When frequency was 50Hz, then the circuit current were 4.71 A,

But when circuit frequency increased from 50Hz to 60Hz, then the current increases as well from 4.71 A to 5.65 A.

Hence proved,

In a capacitive circuit, when frequency increases, the circuit current also increases and vice versa.

f ∝ I

In oral or verbal,

C 1 / XC   and   C ∝  1 / f

I C      and      I 1 / XC   and   I ∝ 1 / Z

Z XC

I  f

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