Math Problem Statement

A series AC circuit has:

Resistor (R=10 ΩR = 10 , \Omega), Inductor (L=0.2 HL = 0.2 , \text{H}), and Capacitor (C=50 μFC = 50 , \mu\text{F}).

AC source with Vpeak=100 VV_\text{peak} = 100 , \text{V} and frequency f=50 Hzf = 50 , \text{Hz}.

Find inductive reactance (XLX_L) and capacitive reactance (XCX_C).

Calculate total impedance (ZZ).

Determine peak current (IpeakI_\text{peak}).

Find phase angle (ϕ\phi).

Is the circuit inductive, capacitive, or resistive?

Solution

Let's go step by step and solve each part of the problem systematically.

1. Inductive Reactance (XLX_L)

Inductive reactance is given by the formula:

XL=2πfLX_L = 2\pi f L

Where:

  • f=50Hzf = 50 \, \text{Hz} (frequency),
  • L=0.2HL = 0.2 \, \text{H} (inductance).

Substituting the given values:

XL=2π×50×0.2X_L = 2\pi \times 50 \times 0.2

XL62.83ΩX_L \approx 62.83 \, \Omega

2. Capacitive Reactance (XCX_C)

Capacitive reactance is given by the formula:

XC=12πfCX_C = \frac{1}{2\pi f C}

Where:

  • f=50Hzf = 50 \, \text{Hz} (frequency),
  • C=50μF=50×106FC = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} (capacitance).

Substituting the given values:

XC=12π×50×50×106X_C = \frac{1}{2\pi \times 50 \times 50 \times 10^{-6}}

XC63.66ΩX_C \approx 63.66 \, \Omega

3. Total Impedance (ZZ)

The total impedance ZZ for a series AC circuit is given by:

Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Where:

  • R=10ΩR = 10 \, \Omega (resistance),
  • XL62.83ΩX_L \approx 62.83 \, \Omega (inductive reactance),
  • XC63.66ΩX_C \approx 63.66 \, \Omega (capacitive reactance).

Now, calculate the difference between the inductive and capacitive reactance:

XLXC=62.8363.66=0.83ΩX_L - X_C = 62.83 - 63.66 = -0.83 \, \Omega

Now, substitute into the impedance formula:

Z=102+(0.83)2Z = \sqrt{10^2 + (-0.83)^2}

Z=100+0.6889Z = \sqrt{100 + 0.6889}

Z100.6889Z \approx \sqrt{100.6889}

Z10.03ΩZ \approx 10.03 \, \Omega

4. Peak Current (IpeakI_{\text{peak}})

The peak current IpeakI_{\text{peak}} is related to the peak voltage and total impedance by:

Ipeak=VpeakZI_{\text{peak}} = \frac{V_{\text{peak}}}{Z}

Where:

  • Vpeak=100VV_{\text{peak}} = 100 \, \text{V} (peak voltage),
  • Z10.03ΩZ \approx 10.03 \, \Omega (total impedance).

Substitute the values:

Ipeak=10010.03I_{\text{peak}} = \frac{100}{10.03}

Ipeak9.97AI_{\text{peak}} \approx 9.97 \, \text{A}

5. Phase Angle (ϕ\phi)

The phase angle ϕ\phi is given by:

ϕ=tan1(XLXCR)\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)

Substitute the values:

ϕ=tan1(0.8310)\phi = \tan^{-1}\left(\frac{-0.83}{10}\right)

ϕ=tan1(0.083)\phi = \tan^{-1}(-0.083)

ϕ4.74\phi \approx -4.74^\circ

6. Inductive, Capacitive, or Resistive Circuit?

To determine whether the circuit is inductive, capacitive, or resistive, we look at the net reactance:

  • If XL>XCX_L > X_C, the circuit is inductive.
  • If XC>XLX_C > X_L, the circuit is capacitive.
  • If XL=XCX_L = X_C, the circuit is purely resistive (with no net reactance).

Since XC>XLX_C > X_L (63.66 Ω > 62.83 Ω), the circuit is capacitive.

Summary of Results:

  1. Inductive Reactance: XL62.83ΩX_L \approx 62.83 \, \Omega
  2. Capacitive Reactance: XC63.66ΩX_C \approx 63.66 \, \Omega
  3. Total Impedance: Z10.03ΩZ \approx 10.03 \, \Omega
  4. Peak Current: Ipeak9.97AI_{\text{peak}} \approx 9.97 \, \text{A}
  5. Phase Angle: ϕ4.74\phi \approx -4.74^\circ
  6. Circuit Type: Capacitive

Would you like further explanations or have any additional questions?

Related Questions:

  1. How does frequency affect the impedance of a series AC circuit?
  2. What happens if the resistance RR is increased in this circuit?
  3. How can we calculate the power factor from the phase angle?
  4. How do inductive and capacitive reactances affect the phase angle?
  5. What is the difference between peak and RMS values for current and voltage in AC circuits?

Tip: In AC circuits, when the impedance is nearly equal to the resistance (ZRZ \approx R), the circuit behaves more resistively, and the current is almost in phase with the voltage.

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Math Problem Analysis

Mathematical Concepts

AC Circuits
Impedance
Reactance
Trigonometry

Formulas

X_L = 2πfL
X_C = 1/(2πfC)
Z = √(R^2 + (X_L - X_C)^2)
I_peak = V_peak / Z
ϕ = tan^(-1)((X_L - X_C) / R)

Theorems

Impedance in AC Circuits
Phase Angle Calculation

Suitable Grade Level

Grades 11-12