Capacitive Reactance Formula: Unit, Frequency & Solved Examples

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Capacitive reactance (X_c) is the opposition offered by a capacitor to the flow of alternating current (AC) in a circuit. It is measured in ohms (Ω). It is inversely proportional to the frequency (f) of the AC signal. The higher the frequency, the lower the capacitive reactance.

Capacitive Reactance measures a capacitor's resistance to alternating current (AC).
Capacitive reactance influences the flow of current and determines the overall impedance of the circuit.
It can also cause a phase shift between the current and voltage waveforms.
Capacitive reactance is essential for AC circuit analysis and design.

Capacitive reactance Formula: Xc = 1/(2πfC), where f is frequency and C is capacitance.

Table of Content

Capacitive Reactance Formula
Capacitive Reactance Concept
Dimensional Formula of Capacitive Reactance
Previous Year Questions
Things to Remember
Sample Questions

Key Terms: Capacitor, Resistance, Current, Voltage, DC circuit, AC circuit, Capacitive reactance, Electrical signal, Resistive circuit

Capacitive Reactance Formula

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The formula to calculate the capacitive reactance is stated below.

X_C = 1/\(\omega\)C = \(\frac{1}{2 \pi\;f\;C}\)

where,

X_C = Capacitive reactance
\(\omega\) = Angular frequency
C = Capacitance
f = Frequency

Capacitive Reactance Formula Solved Examples

Example 1 – A capacitor has a capacitance of 10 µF and is connected to an AC circuit with a frequency of 50 Hz. What is the capacitive reactance of the capacitor?

Solution: Xc = 1 / (2πfC)

Xc = 1 / (2π * 50 * 10 * 10^-6)

Xc = 318.31 ohms (Ω)

Example 2 – A radio circuit uses a capacitor with a capacitance of 100 µF. The radio receives a signal with a frequency of 1 MHz. What is the capacitive reactance of the capacitor?

Solution: Given:

Capacitance (C) = 100 µF = 100 × 10⁻⁶ F

Frequency (f) = 1 MHz = 1 × 10⁶ Hz

Capacitive reactance (Xc) can be calculated using the formula:

Xc = 1/(2πfC)

Substituting these values in the formula:

Xc = 1 / (2π * 1 × 10⁶ * 100 × 10⁻⁶) ≈ 159.15 Ω

Therefore, the capacitive reactance of the capacitor is approximately 159.15 ohms.

Capacitive Reactance

[Click Here for Sample Questions]

A capacitor is an electronic component that stores electrical energy.

In a DC circuit, a capacitor charges until it reaches full capacity and then effectively blocks current flow.
In an AC circuit, a capacitor allows current to flow but opposes it, a phenomenon known as capacitive reactance.
In a purely capacitive circuit, capacitive reactance acts like resistance in a resistive circuit, limiting the current flow.
Capacitive Reactance in Series is less than the capacitance of any single capacitor.
The formula for total capacitance in series is: C_total = 1/(1/C₁ + 1/C₂ + ... + 1/C_in)
The total capacitance of capacitors connected in parallel is equal to the sum of the individual capacitors' values.
The formula for total capacitance in parallel is: C_total = C₁ + C₂ + ... + C_in

Capacitive Reactance

Capacitive Reactance

Capacitive reactance is inversely proportional to both capacitance and frequency.

As f increases, the capacitor passes more charge, resulting in a greater current flow and a lower internal impedance.
A capacitor connected to a circuit that changes over a range of frequencies is frequency-dependent.

Capacitance depends on Frequency

Capacitance depends on the Frequency

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Dimensional Formula of Capacitive Reactance

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To calculate the dimensional formula of capacitive reactance, we need the capacitive reactance of frequency and capacitance.

Dimensional formula of f = [T^-1]

Now, capacitance C = qV

Dimensional formula of charge q = [A¹T¹]
Dimensional formula of voltage V = dimensional formula of energy/ dimensional formula of charge
Dimensional formula of voltage V = [M¹L²T^-2] / [A¹T¹] = [M¹L²T^-3A^-1]
Dimensional formula of capacitance C = [A¹T¹] / [M¹L²T^-3A^-1] = [M^-1L^-2T⁴A²]

Dimensional formula of capacitive reactance = 1/(dimensional formula of frequency) x (dimensional formula of capacitance)

Dimensional formula of capacitive reactance = 1 / [T^-1]⋅[M^-1L^-2T⁴A²]

→ 1 / [M^-1L^-2T³A²]

→ [M¹L²T^-3A^-2]

So, the dimensional formula of capacitive reactance is [M¹L²T^-3A^-2].

Previous Year Questions

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Things to Remember

The opposition faced by the current in an AC circuit because of the capacitor is called capacitive reactance.
Capacitive reactance is represented by X_C and its SI unit is ohm (\(\Omega\)).
The formula for capacitive reactance is Xc = 1/(2πfC), where C is the capacitance of the capacitor.
The dimensional formula of capacitive reactance is [M¹L²T^-3A^-2].
As the frequency of the AC signal increases, the capacitive reactance decreases.
As the capacitance of the capacitor increases, the capacitive reactance decreases.
The total capacitance of capacitors connected in series is less than the capacitance of any single capacitor.
The total capacitance of capacitors connected in parallel is equal to the sum of the individual capacitors' values.

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Sample Questions

Ques 1: Calculate the capacitive reactance if a 40 mF capacitor is connected to a voltage source generating a 50 Hz signal. (2 marks)

Ans: C = 40 mF

f = 50 Hz

The formula to calculate capacitive reactance,

X_C = 1/2\(\Pi\)fC

1/2 × 3.14 × 50 × 40 = 0.0796 \(\Omega\)

So, the capacitive reactance is 0.0796 \(\Omega\).

Ques 2: Calculate the capacitance of a capacitor in an AC circuit that has a capacitive reactance of 1.5 × 10^-3\(\Omega\) and the input signal has a frequency of 100 Hz. (2 marks)

Ans: X_C = 1.5 × 10^-3\(\Omega\)

f = 100 Hz

To calculate the capacitance, we have

X_C =1/2\(\Pi\)fC

1.5 × 10^-3 = 12× 3.14 × 100 × C

C = 1/2 × 3.14 × 100 × 1.5 = 1.0616 F = 1061.6 mF

So, the capacitance of the capacitor is 1061.6 mF.

Ques 3: Calculate the capacitive reactance of a capacitor whose capacitance is 220 nF at frequencies of 1 kHz and 20 kHz. (2 marks)

Ans: C = 220 nF = 220 × 10^-9 F

Formula for capacitive reactance, X_C =1/2\(\Pi\)fC

For 1 kHz frequency,

X_C =1/2 × 3.14 × 1000 × 220×10⁹ = 723.8 \(\Omega\)

For 20 kHz frequency,

X_C =1/2 × 3.14 × 20 × 1000 × 220×10⁹ = 36.19 \(\Omega\)

So, when the frequency increases from 1 kHz to 20 kHz, the capacitive reactance decreases from 723.8 \(\Omega\) to 36.19 \(\Omega\).

Ques 4: Calculate at what frequency the capacitive reactance of a capacitor of 2.2 µF will be 20 \(\Omega\). (3 marks)

Ans: C = 2.2 µF = 2.2× 10^-6 F

X_C = 20 \(\Omega\)

Capacitive reactance X_C =1/2\(\Pi\)fC

20 = 1/2 × 3.14 × f × 2.2×10⁶

f = 1/2 × 3.14 × 2 × 2.2×10⁵ = 3619 Hz

So, the frequency at that given time will be 3619 Hz.

Ques 5: Calculate the capacitance of a capacitor whose capacitive reactance is 200 ohms and is connected to a voltage supply with a frequency of 50 Hz. (3 marks)

Ans: X_C = 200 \(\Omega\)

f = 50 Hz

Calculating capacitance,

X_C =1/2\(\Pi\)fC

200 = 1/2 × 3.14 × 50 × C

C = 1.592 × 10^-5 F = 15.92 µF

So, the capacitance of the capacitor is 15.92 µF.

Ques 6: Determine the capacitive reactance of a capacitor connected in a DC circuit. (3 marks)

Ans: Now, in a DC circuit, the voltage supply is constant. Hence, the frequency is zero.

f = 0

From the formula for capacitive reactance, we know that capacitive reactance is inversely proportional to the frequency of the voltage supply.

X_C ∝ 1f

So, when the frequency is zero, the capacitive reactance will be infinite.

X_C = ∞

So, when a capacitor is connected to a DC source, its capacitive reactance is infinite.

Ques 7: Three capacitors, 12 µF, 20 µF and 30 µF are connected to a voltage source with a frequency of 60 Hz. Calculate the capacitive reactance when they are connected in series and parallel connections. (5 marks)

Ans: C₁ = 12 µF

C₂ = 20 µF

C₃ = 30 µF

f = 60 Hz

For series connection

\(\text{Series Capacitances}\)

\(C_{total}= \frac{1}{\frac{1}{C_{1}} + \frac{1}{C_{2}} + ... \frac{1}{C_{n}}}\)

1/C₁ = 112 = 0.0833 µF

1/C₂ = 120 = 0.05 µF

1/C₃ = 130 = 0.0333 µF

C_total = 10.0833 + 0.05 + 0.0333= 10.1666= 6 µF

To calculate capacitive reactance,

X_C =1/2\(\Pi\)fC

X_C = 1/2 × 3.14 × 60 × 6×106 = 442.32 \(\Omega\)

For parallel connection

Resultant capacitance, C_total = C₁ + C₂ + … + C_n

Here, C_total = C₁ + C₂ + C₃

C_total = 12 + 20 + 30 = 62 µF

To calculate capacitive reactance,

X_C =1/2\(\Pi\)fC

X_C = 1/2 × 3.14 × 60 × 62×10⁶ = 42.805 \(\Omega\)

So, the capacitive reactance of this circuit when connected in series and parallel connections is 442.32 \(\Omega\) and 42.805 \(\Omega\), respectively.

Ques 8: Define the dielectric constant of a medium. What is its unit? [CBSE 2011] (2 marks)

Ans. Dielectric When a dielectric slab is introduced between the plates of a charged capacitor or in the region of the electric field, an electric field E_P is induced inside the dielectric due to induced charge on the dielectric in a direction opposite to the direction of the applied external electric field. Hence, the net electric field inside the dielectric gets reduced to E₀ – E_P, where E₀ is the external electric field. The ratio of applied external electric field and reduced electric field is known as the dielectric constant K of the dielectric medium, i.e.

\(K=\frac{E_0}{E_0 - E_{p}}\)

Ques 9: Explain in detail about the Equipotential surface. (3 marks)

Ans. Equipotential Surface A surface that has the same electrostatic potential at every point on it is known as an equipotential surface.

The shape of the equipotential surface due to

The line charge is cylindrical.

A point charge is spherical as shown alongside:

Equipotential surfaces do not intersect each other as it gives two directions of electric field E at an intersecting point which is not possible.
Equipotential surfaces are closely spaced in the region of a strong electric field and vice-versa.
The electric field is always normal to the equipotential surface at every point of it and directed from one equipotential surface at the higher potential to the equipotential surface at the lower potential.
Work done in moving a test charge from one point of equipotential surface to another is zero.

Equipotential Surface

Ques 10: What is Electrostatic shielding? (3 marks)

Ans. Electrostatic Shielding is the process that involves the making of a region free from any electric field known as electrostatic shielding.

Electrostatic Shielding

It happens due to the fact that no electric field exists inside a charged hollow conductor. The potential inside a shell is constant. In this way, we can also conclude that the field inside the shell (hollow conductor) will be zero.

Ques 11: Is capacitive reactance AC or DC? (3 marks)

Ans. Capacitive reactance (Xc) is an AC property. It is a measure of a capacitor's opposition to alternating current (AC). DC circuits do not have a frequency, so the concept of capacitive reactance does not apply to them.

In an AC circuit, the capacitive reactance of a capacitor is determined by its capacitance (C) and the frequency (f) of the AC signal. The formula for capacitive reactance is:

Xc = 1 / (2πfC)
where:

Xc is the capacitive reactance in ohms (Ω)
C is the capacitance in farads (F)
f is the frequency in hertz (Hz)

As the frequency of the AC signal increases, the capacitive reactance decreases. This means that the capacitor will oppose the AC signal less at higher frequencies.

As the capacitance of the capacitor increases, the capacitive reactance also decreases. This means that a larger capacitor will oppose the AC signal less than a smaller capacitor.

Ques 12: What are XL and XC in RLC circuits? (5 marks)

Ans. XL and XC are respectively the inductive reactance and capacitive reactance in RLC circuits. These are quantities that represent the opposition of inductors and capacitors to alternating current (AC), respectively.

Inductive reactance (XL) is directly proportional to the frequency of the AC signal and the inductance (L) of the inductor. It is measured in ohms (Ω) and can be calculated using the formula:

XL = 2πfL

Capacitive reactance (XC) is inversely proportional to the frequency of the AC signal and the capacitance (C) of the capacitor. It is also measured in ohms (Ω) and can be calculated using the formula:

XC = 1 / (2πfC)

In an RLC circuit, the total impedance (Z) is the combination of the resistance (R), inductive reactance (XL), and capacitive reactance (XC). It is calculated using the formula:

Z = √(R² + (XL - XC)²)

The impedance of a circuit determines the amount of current that will flow through it when an AC voltage is applied.

Ques 13: A power supply circuit uses a capacitor with a capacitance of 470 µF to filter out high-frequency noise. The supply voltage has a frequency of 60 Hz. What is the capacitive reactance of the capacitor at this frequency? (5 marks)

Ans. Given:

Capacitance (C) = 470 µF = 470 × 10⁻⁶ F

Frequency (f) = 60 Hz

Capacitive reactance (Xc) can be calculated using the formula:

Xc = 1 / (2πfC)

Plugging in the values:

Xc = 1 / (2π * 60 * 470 × 10⁻⁶) ≈ 54.01 Ω

Therefore, the capacitive reactance of the capacitor at 60 Hz is approximately 54.01 ohms.

Ques 14: A lighting circuit uses a capacitor with a capacitance of 220 µF to prevent flickering caused by fluctuations in the AC power supply. The power supply has a frequency of 50 Hz. What is the capacitive reactance of the capacitor at this frequency? (3 marks)

Ans. Given:

Capacitance (C) = 220 µF = 220 × 10⁻⁶ F

Frequency (f) = 50 Hz

Capacitive reactance (Xc) can be calculated using the formula:

Xc = 1 / (2πfC)

Plugging in the values:

Xc = 1 / (2π * 50 * 220 × 10⁻⁶) ≈ 723.89 Ω

Therefore, the capacitive reactance of the capacitor at 50 Hz is approximately 723.89 ohms.

Ques 15: Give the derivation for capacitive reactance. (5 marks)

Ans. The derivation for capacitive reactance starts with the basic definition of a capacitor. A capacitor is an electrical device that stores energy in an electric field. The amount of energy stored in a capacitor is given by the formula:

U = (1/2)CV²

where:

U is the energy stored in joules (J)
C is the capacitance in farads (F)
V is the voltage across the capacitor in volts (V)

The current through a capacitor is given by the formula:

I = dQ/dt

where:

I is the current in amperes (A)
Q is the charge on the capacitor in coulombs (C)
t is time in seconds (s)

The charge on a capacitor is proportional to the voltage across the capacitor:

Q = CV
Substituting this equation into the formula for current, we get:

I = C(dV/dt)

This equation shows that the current through a capacitor is proportional to the rate of change of the voltage across the capacitor.

The impedance of a capacitor is a measure of its opposition to AC current flow. It is defined as the ratio of the voltage across the capacitor to the current through the capacitor:

Z = V/I

Substituting the formula for current into this equation, we get:

Z = V / (CdV/dt)
This equation shows that the impedance of a capacitor is inversely proportional to the rate of change of the voltage across the capacitor.

The capacitive reactance of a capacitor is the imaginary part of its impedance. It is denoted by the symbol Xc and is calculated using the formula:

Xc = 1 / (2πfC)

where:

Xc is the capacitive reactance in ohms (Ω)
f is the frequency of the AC signal in hertz (Hz)
C is the capacitance in farads (F)

This equation shows that the capacitive reactance of a capacitor is inversely proportional to the frequency of the AC signal and the capacitance of the capacitor.

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Capacitive Reactance Formula: Unit, Frequency & Solved Examples

Capacitive Reactance Formula

Capacitive Reactance Formula Solved Examples

Capacitive Reactance

Dimensional Formula of Capacitive Reactance

Previous Year Questions

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

2.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

5.
A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true?

6.
An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

Similar Physics Concepts

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Capacitive Reactance Formula: Unit, Frequency & Solved Examples

Capacitive Reactance Formula

Capacitive Reactance Formula Solved Examples

Capacitive Reactance

Dimensional Formula of Capacitive Reactance

Previous Year Questions

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

2.The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

3.The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

5.A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true?

6.An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

3.
The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

5.
A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true?

6.
An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.