# 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**“.

*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 a capacitive circuit:

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

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__ “X_{C}” are inversely proportional to each others.

Related Questions:

- Which Transformer is More Efficient When Operates on 50Hz or 60Hz?
- Can We Operate a 60Hz Transformer on 50Hz Supply Source and Vice Versa?
- Is it Possible to Operate a 50Hz Transformer on 5Hz or 500Hz Frequency?

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:

- Voltage = V = 3000 V
- Capacitance = C = 5 µF
- Frequency =
*f*= 50 Hz

To find the capacitive reactance;

X_{C} = 1 / 2π*f*C

X_{C} = 1 / (2 x 3.1415 x 50 x 5×10^{-6})

X_{C} = 636.94 Ω

Now, current in the capacitive circuit:

I = V / X_{C}

I = 3000 V / 636.94 Ω

** I = 4.71 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, C = 5 µ Farads, *f* = 60 Hz.

X_{C} = 1 / 2π*f*C = 1 / (2 x 3.1415 x 60 x 5×10^{-6}) = 530.63Ω

I = V / X_{C} = 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,

- Capacitive reactance is a kind of resistance. When resistance increases, the circuit current decreases and vice versa.
- Capacitance is inversely proportional to the capacitive reactance and frequency.

**C ****∝**** 1 / X _{C} and C ∝ 1 / f**

- Current is directly proportional to the capacitance and inversely proportional to capacitive reactance and impedance.

**I ****∝**** C and I ****∝**** 1 / X _{C} and I ∝ 1 / Z**

- Impedance is directly proportional to the capacitive reactance

**Z ****∝**** X _{C}**

**In a capacitive circuit, Frequency is directly proportional to the current.**

**I ****∝** * f*

Related Questions/Answers:

- According to Ohm’s Law, I ∝ V, But I ∝ 1/V in Power Equation. How do you Explain?
- 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)? - Which One is More Dangerous? 50Hz or 60Hz in 120V/230V & Why?
- Why Can’t a 12V Car Battery Electrocute You?
- Which One is More Dangerous? 120V or 230V and Why?

I want the real answers based on the above question