AC Circuit Analysis: U=110V, F=50Hz, L=0.04H, R=9Ω, C=200µF

by TextBrain Team 60 views

Alright, guys, let's dive into analyzing this AC circuit! We've got a circuit with a voltage source of 110V at a frequency of 50Hz, an inductor of 0.04 Henrys, a resistor of 9 ohms, and a capacitor of 200 microfarads. Buckle up; we're about to crunch some numbers and understand how this circuit behaves.

Understanding the Components

Before we start calculating, let's briefly review what each component does in an AC circuit.

  • Resistor (R): The resistor opposes the flow of current and dissipates energy in the form of heat. The resistance is constant regardless of the frequency of the AC source.
  • Inductor (L): The inductor opposes changes in current. It stores energy in a magnetic field. Its opposition to current flow, called inductive reactance, increases with frequency.
  • Capacitor (C): The capacitor opposes changes in voltage. It stores energy in an electric field. Its opposition to current flow, called capacitive reactance, decreases with frequency.

Calculating Reactances

The first step is to calculate the inductive reactance (XL) and capacitive reactance (XC).

Inductive Reactance (XL)

The inductive reactance is given by the formula:

XL = 2 * π * F * L

Where:

  • XL is the inductive reactance in ohms.
  • π (pi) is approximately 3.14159.
  • F is the frequency in Hertz.
  • L is the inductance in Henrys.

Plugging in the values:

XL = 2 * 3.14159 * 50 * 0.04

XL ≈ 12.57 ohms

So, the inductive reactance is approximately 12.57 ohms. This means the inductor opposes the flow of AC current with an effective resistance of 12.57 ohms at 50 Hz.

Capacitive Reactance (XC)

The capacitive reactance is given by the formula:

XC = 1 / (2 * π * F * C)

Where:

  • XC is the capacitive reactance in ohms.
  • π (pi) is approximately 3.14159.
  • F is the frequency in Hertz.
  • C is the capacitance in Farads. Important: We need to convert microfarads to farads. 200 μF = 200 * 10^-6 F = 0.0002 F

Plugging in the values:

XC = 1 / (2 * 3.14159 * 50 * 0.0002)

XC ≈ 15.92 ohms

Therefore, the capacitive reactance is approximately 15.92 ohms. The capacitor opposes the flow of AC current with an effective resistance of 15.92 ohms at 50 Hz.

Calculating Impedance (Z)

Now that we have the resistance and the reactances, we can calculate the total impedance (Z) of the circuit. Impedance is the total opposition to current flow in an AC circuit, considering both resistance and reactance. Because the resistor, inductor, and capacitor are in series, the impedance is calculated as:

Z = √(R² + (XL - XC)²)

Where:

  • Z is the impedance in ohms.
  • R is the resistance in ohms.
  • XL is the inductive reactance in ohms.
  • XC is the capacitive reactance in ohms.

Plugging in the values:

Z = √(9² + (12.57 - 15.92)²)

Z = √(81 + (-3.35)²)

Z = √(81 + 11.22)

Z = √92.22

Z ≈ 9.60 ohms

The total impedance of the circuit is approximately 9.60 ohms. This value is crucial because it determines how much current will flow through the circuit for a given voltage.

Calculating Current (I)

Using Ohm's Law for AC circuits, we can calculate the current (I) flowing through the circuit:

I = U / Z

Where:

  • I is the current in Amperes.
  • U is the voltage in Volts.
  • Z is the impedance in ohms.

Plugging in the values:

I = 110 / 9.60

I ≈ 11.46 Amperes

The current flowing through the circuit is approximately 11.46 Amperes. This is the RMS (root mean square) value of the current. It represents the effective current that delivers the same power as a DC current of the same value.

Phase Angle (Φ)

The phase angle (Φ) tells us the phase difference between the voltage and the current in the circuit. It is calculated as:

Φ = arctan((XL - XC) / R)

Where:

  • Φ is the phase angle in radians or degrees.
  • XL is the inductive reactance in ohms.
  • XC is the capacitive reactance in ohms.
  • R is the resistance in ohms.

Plugging in the values:

Φ = arctan((12.57 - 15.92) / 9)

Φ = arctan(-3.35 / 9)

Φ = arctan(-0.372)

Φ ≈ -20.4 degrees

The phase angle is approximately -20.4 degrees. The negative sign indicates that the current leads the voltage. This is because the capacitive reactance is greater than the inductive reactance, making the circuit predominantly capacitive.

Power Calculations

Now, let's calculate the power in the circuit.

Apparent Power (S)

The apparent power is the product of the voltage and current, without considering the phase angle:

S = U * I

Where:

  • S is the apparent power in Volt-Amperes (VA).
  • U is the voltage in Volts.
  • I is the current in Amperes.

Plugging in the values:

S = 110 * 11.46

S ≈ 1260.6 VA

The apparent power is approximately 1260.6 VA.

Active Power (P)

The active power (also called real power) is the power actually consumed by the circuit and dissipated as heat in the resistor:

P = U * I * cos(Φ)

Where:

  • P is the active power in Watts (W).
  • U is the voltage in Volts.
  • I is the current in Amperes.
  • Φ is the phase angle.

Plugging in the values:

P = 110 * 11.46 * cos(-20.4°)

P = 1260.6 * 0.937

P ≈ 1181.2 Watts

The active power is approximately 1181.2 Watts. This is the power that is actually used by the circuit to do work (in this case, dissipated as heat in the resistor).

Reactive Power (Q)

The reactive power is the power exchanged between the inductor and capacitor. It is not consumed but oscillates between these components and the source.

Q = U * I * sin(Φ)

Where:

  • Q is the reactive power in Volt-Ampere Reactive (VAR).
  • U is the voltage in Volts.
  • I is the current in Amperes.
  • Φ is the phase angle.

Plugging in the values:

Q = 110 * 11.46 * sin(-20.4°)

Q = 1260.6 * (-0.348)

Q ≈ -438.7 VAR

The reactive power is approximately -438.7 VAR. The negative sign indicates that the circuit is predominantly capacitive, meaning it supplies reactive power.

Power Factor (PF)

The power factor is the ratio of active power to apparent power:

PF = P / S

Or, equivalently:

PF = cos(Φ)

In this case:

PF = 1181.2 / 1260.6

PF ≈ 0.937

Or:

PF = cos(-20.4°)

PF ≈ 0.937

The power factor is approximately 0.937. A power factor close to 1 indicates that the circuit is efficient in using power. A lower power factor means that a larger portion of the apparent power is reactive and does not contribute to useful work. In this case, it is a leading power factor because the current leads the voltage.

Summary

Let's recap our findings for this AC circuit:

  • Inductive Reactance (XL): 12.57 ohms
  • Capacitive Reactance (XC): 15.92 ohms
  • Impedance (Z): 9.60 ohms
  • Current (I): 11.46 Amperes
  • Phase Angle (Φ): -20.4 degrees (current leads voltage)
  • Apparent Power (S): 1260.6 VA
  • Active Power (P): 1181.2 Watts
  • Reactive Power (Q): -438.7 VAR
  • Power Factor (PF): 0.937 (leading)

Conclusion

By analyzing the given AC circuit, we determined the impedance, current, phase angle, and various power values. The circuit is predominantly capacitive, as indicated by the negative phase angle and the leading power factor. This analysis helps us understand how the circuit behaves under AC conditions and allows us to optimize its performance for specific applications. Hope this detailed analysis helps you understand the intricacies of this AC circuit! Let me know if you have any other questions, guys!