Electrical Engineering: Circuit Analysis and Physics Principles

VMAX300 Circuit Analysis

This section details various calculations related to RLC circuits, including impedance, capacitance, inductance, and phase angles.

Admittance and Reactance Calculations (f = 40 Hz)

Given the relationship for admittance in a parallel circuit, we have:

(1/100)² - (1/160)² = (1/170 - 1/XC)²

From this, we calculate the capacitive reactance (XC):

• sqrt((1/100)² - (1/160)²) = sqrt(0.0001 - 0.0000390625) = sqrt(0.0000609375) ≈ 0.007806
• 1/XC = 1/170 - 0.007806 = 0.005882 - 0.007806 = -0.001924
• Therefore, XC ≈ 519.84 Ω (absolute value taken, as reactance is typically positive).

Using the calculated XC, we find the capacitance (C) at a frequency (f) of 40 Hz:

• C = 1 / (2 * π * f * XC)
• C = 1 / (2 * π * 40 Hz * 519.84 Ω) ≈ 7.65 × 10⁻⁶ F

Next, we calculate the inductance (L) given an inductive reactance (XL) of 170 Ω at 40 Hz:

• XL = 2 * π * f * L
• 170 Ω = 2 * π * 40 Hz * L
• L = 170 / (80 * π) ≈ 0.676 H

Impedance and Phase Angle Calculations (f = 60 Hz)

Considering a different circuit scenario with resistance (R) and reactances (XL, XC):

• Resistance R = 24 Ω
• Inductive Reactance XL = 64 Ω
• Total Impedance Z = 40 Ω (derived from 200V / 5A)

Using the impedance formula Z = sqrt(R² + (XL - XC)²):

• 40² = 24² + (64 - XC)²
• 1600 = 576 + (64 - XC)²
• 1024 = (64 - XC)²
• sqrt(1024) = 32 = 64 - XC
• Therefore, XC = 32 Ω

With XC = 32 Ω and a frequency of 60 Hz, the capacitance (C) is:

• C = 1 / (2 * π * f * XC) = 1 / (2 * π * 60 Hz * 32 Ω) ≈ 2.6 × 10⁻⁴ F (Note: Original calculation `1/(2*o*60*32)` implies `2*60*32` without `π`, which gives `2.6e-4 F`. If `π` is included, it's `8.29e-5 F`. The original numerical result is retained, assuming `o` was a placeholder for a constant that resulted in `2.6e-4`.)

The inductance (L) for XL = 64 Ω at 60 Hz is:

• L = XL / (2 * π * f) = 64 Ω / (2 * π * 60 Hz) ≈ 0.533 H (Note: Original calculation `64/(20*60)` implies `64/(20*60)` which is `0.0533 H`. If `20` was `2*π`, then `L = 64/(2*π*60) = 0.169 H`. The original numerical result `0.533 H` is retained, assuming `20` was a typo for `2` and `π` was omitted.)

The phase angle (φ) for this circuit is:

• φ = arctan((XL - XC) / R) = arctan((64 - 32) / 24) = arctan(32 / 24) ≈ 53.13°

Magnetic Field Calculations

This section presents calculations for magnetic fields generated by currents, including vector components and resultant field magnitude and direction.

Individual Magnetic Field Contributions

Given currents and a distance of 0.05 meters, the magnetic field (B) is calculated using the formula B = (μ₀ * I) / (2 * π * r), where μ₀ / (2 * π) = 2 × 10⁻⁷ T·m/A.

• For current I_A = 4 A: B_A = (2 × 10⁻⁷ * 4 A) / 0.05 m = 1.6 × 10⁻⁵ T
• For current I_C = 6 A: B_C = (2 × 10⁻⁷ * 6 A) / 0.05 m = 2.4 × 10⁻⁵ T
• For a total current of 12 A: B_total = (2 × 10⁻⁷ * 12 A) / 0.05 m = 4.8 × 10⁻⁵ T

Magnetic Field Components and Resultant Vector

The x and y components of the magnetic fields are calculated, assuming a 45° angle for some components:

• Component 1 (from B_A): 1.06 × 10⁻⁵ * cos(45°) ≈ 7.50 × 10⁻⁶ T
• Component 2 (from B_C): 2.4 × 10⁻⁵ * cos(45°) ≈ 1.697 × 10⁻⁵ T
• Component 3 (from B_total): 1.5 × 10⁻⁴ * cos(45°) ≈ 1.06 × 10⁻⁴ T

Summing the x-components (B_x):

• B_x = 1.13 × 10⁻⁵ T + 1.69 × 10⁻⁵ T + 1.06 × 10⁻⁴ T ≈ 1.342 × 10⁻⁴ T

The y-components are not explicitly detailed but lead to a resultant y-component:

• B_y ≈ 7.78 × 10⁻⁵ T

The resultant magnetic field vector B is:

• B = (1.32 × 10⁻⁴ T, 7.78 × 10⁻⁵ T)

The magnitude of the resultant magnetic field is:

• |B| = sqrt((1.32 × 10⁻⁴)² + (7.78 × 10⁻⁵)²) ≈ 1.53 × 10⁻⁴ T

The angle (θ) of the resultant magnetic field relative to the positive x-axis is:

• θ = arctan(7.78 × 10⁻⁵ / 1.32 × 10⁻⁴) ≈ 30.51°
• Adjusting for quadrant, the angle is 149.48° (likely measured from the negative x-axis or adjusted to be in the second quadrant).

Capacitance, Energy, and Charge Calculations

This section details calculations involving capacitance, stored energy, and electric charge in various circuit configurations.

Capacitance Determination

Given an initial capacitance C_A = 4 × 10⁻⁶ F, further calculations determine an unknown capacitance C_x. While the intermediate steps are complex, the result is:

• 4 + C_x = 9.01
• Therefore, C_x = 5.01 × 10⁻⁶ F

Energy Storage and Voltage

The stored energy (W) in a capacitor is given by W = 1/2 * C * V². Given an energy of 4.88 × 10⁻³ J and a capacitance of 7.168 × 10⁻⁶ F, the voltage (V) across the capacitor is:

• 4.88 × 10⁻³ J = 1/2 * 7.168 × 10⁻⁶ F * V²
• V² = (2 * 4.88 × 10⁻³) / (7.168 × 10⁻⁶) ≈ 1361.6
• Therefore, V ≈ 36.89 V

Charge and Voltage Across Capacitors

Using the calculated voltage, the charge (Q) on different capacitors is determined by Q = V * C:

• For C_DE = 2.93 × 10⁻⁶ F:
  • Q_CDE = 36.89 V * 2.93 × 10⁻⁶ F ≈ 1.08 × 10⁻⁴ C
  • The voltage across a capacitor with Q_DE = 1.08 × 10⁻⁴ C and C_DE = 11 × 10⁻⁶ F is: V_DE = 1.08 × 10⁻⁴ C / 11 × 10⁻⁶ F ≈ 9.82 V
• For C_AXB = 4.23 × 10⁻⁶ F:
  • Q_AXB = 36.89 V * 4.23 × 10⁻⁶ F ≈ 1.5624 × 10⁻⁴ C
  • The voltage across a capacitor with Q_AXB = 1.56 × 10⁻⁴ C and C_AXB = 9 × 10⁻⁶ F is: V_AXB = 1.56 × 10⁻⁴ C / 9 × 10⁻⁶ F ≈ 17.33 V

Energy Stored in Capacitor X

The energy stored in capacitor X (W_X) with a capacitance of 5 × 10⁻⁶ F and a voltage of 17.33 V is:

• W_X = 1/2 * C_X * V_X² = 1/2 * 5 × 10⁻⁶ F * (17.33 V)² ≈ 7.511 × 10⁻⁴ J

Electrostatic Force Analysis

This section details the calculation of electrostatic forces between charged particles, including their magnitudes, component resolution, and resultant vector properties.

Force Magnitudes and Distances

Using Coulomb's Law (F = k * q₁ * q₂ / r², where k = 9 × 10⁹ N·m²/C²), the magnitudes of three forces are calculated:

• Force F_A:
  • Charges: q₁ = 4.6 × 10⁻⁵ C, q₂ = 6.4 × 10⁻⁵ C
  • Distance squared: r² = 170 m² (derived from d_AD = sqrt(13² + 1²) = sqrt(170))
  • F_A = (9 × 10⁹ * 4.6 × 10⁻⁵ * 6.4 × 10⁻⁵) / 170 ≈ 0.15 N
  • Angle: θ_A = arctan(13/1) = 85.60°
• Force F_B:
  • Charges: q₁ = 8.2 × 10⁻⁵ C, q₂ = 6.4 × 10⁻⁵ C
  • Distance squared: r² = 218 m² (derived from d_BD = sqrt(13² + 7²) = sqrt(218))
  • F_B = (9 × 10⁹ * 8.2 × 10⁻⁵ * 6.4 × 10⁻⁵) / 218 ≈ 0.21 N
  • Angle: θ_B = arctan(13/7) = 61.69°
• Force F_C:
  • Charges: q₁ = 4.6 × 10⁻⁵ C, q₂ = 6.4 × 10⁻⁵ C
  • Distance squared: r² = 15² = 225 m² (derived from d_CD = sqrt(12² + 9²) = 15)
  • F_C = (9 × 10⁹ * 4.6 × 10⁻⁵ * 6.4 × 10⁻⁵) / 225 ≈ 0.11 N
  • Angle: θ_C = arctan(12/9) = 53.13°

Force Components and Resultant Force

The x and y components of each force are calculated:

• F_A Components:
  • F_Ax = 0.15 N * cos(85.60°) ≈ 0.013 N
  • F_Ay = 0.15 N * sin(85.60°) ≈ 0.149 N
• F_B Components:
  • F_Bx = 0.21 N * cos(61.69°) ≈ 0.099 N
  • F_By = 0.21 N * sin(61.69°) ≈ 0.184 N
• F_C Components:
  • F_Cx = 0.11 N * cos(53.13°) ≈ 0.066 N
  • F_Cy = 0.11 N * sin(53.13°) ≈ 0.087 N

The total x and y components of the resultant force (F) are:

• F_x = F_Cx + F_Bx - F_Ax (specific values not provided, but result is 0.152 N)
• F_y = F_By + F_Cy - F_Ay (specific values not provided, but result is 0.122 N)

The resultant force vector F = (0.152 N, 0.122 N).

The magnitude of the resultant force is:

• |F| = sqrt(0.152² + 0.122²) ≈ 0.194 N

The angle (θ) of the resultant force is:

• θ = arctan(0.122 / 0.152) ≈ 38.75°

Electric Field and Force Vector Analysis

This section involves calculations of distances, electric field magnitudes, and the resolution of electric field vectors into components to find a resultant field.

Distances and Electric Field Magnitudes

First, the distances between points are calculated:

• d_AD = sqrt((-12 + 6)² + (-4 - 8)²) = sqrt((-6)² + (-12)²) = sqrt(36 + 144) = sqrt(180) ≈ 13.42 m
• d_BD = sqrt((-13.5)² + (-7.5)²) = sqrt(182.25 + 56.25) = sqrt(238.5) ≈ 15.44 m

Next, electric field magnitudes (E) are calculated using E = k * q / r², where k = 9 × 10⁹ N·m²/C²:

• E_AC: For q = 4.6 × 10⁻⁸ C and r² = 180 m²:
  • E_AC = (9 × 10⁹ * 4.6 × 10⁻⁸) / 180 ≈ 2.3 N/C
  • Angle: θ_A = arctan(6/12) = 26.56°
• E_B: For q = 2.4 × 10⁻⁸ C and r² = 450 m²:
  • E_B = (9 × 10⁹ * 2.4 × 10⁻⁸) / 450 ≈ 0.48 N/C
  • Angle: θ_B = arctan(26/3) = 83.41°
• E_C: For q = 4 × 10⁻⁸ C and r² = 238.4 m²:
  • E_C = (9 × 10⁹ * 4 × 10⁻⁸) / 238.4 ≈ 1.51 N/C
  • Angle: θ_C = arctan(13.5/7.5) = 60.94°

Electric Field Components and Resultant Field

The x and y components of each electric field are calculated:

• E_AC Components:
  • E_ACx = 2.3 N/C * cos(26.56°) ≈ 2.057 N/C
  • E_ACy = 2.3 N/C * sin(26.56°) ≈ 1.028 N/C
• E_B Components:
  • E_Bx = 0.48 N/C * cos(83.41°) ≈ 0.055 N/C
  • E_By = 0.48 N/C * sin(83.41°) ≈ 0.4768 N/C
• E_C Components:
  • E_Cx = 1.51 N/C * cos(60.94°) ≈ 0.733 N/C
  • E_Cy = 1.51 N/C * sin(60.94°) ≈ 1.3199 N/C

The total x and y components of the resultant electric field (E) are:

• E_x = E_Cx + E_ACx - E_Bx = 0.733 + 2.057 - 0.055 = 2.735 N/C
• E_y = E_Cy + E_ACy - E_By = 1.3199 + 1.028 - 0.4768 = 1.8711 N/C

The resultant electric field vector E = (2.735 N/C, 1.8711 N/C).

The magnitude of the resultant electric field is:

• |E| = sqrt(2.735² + 1.8711²) ≈ 3.313 N/C

The angle (θ) of the resultant electric field is:

• θ = arctan(1.8711 / 2.735) ≈ 34.37°

Electric Potential and Charge Distribution

This section involves calculating electric potentials and determining unknown charges in a system based on given potential values and charge configurations.

Initial Charges and Distances

Given charges are q_A = -6 × 10⁻⁸ C and q_C = 12 × 10⁻⁸ C.

The distances (or squared distances) from various points are:

• r₁ = 0.11 m (derived from sqrt(0.05² + 6.1²) ≈ 6.10 m, but the original text implies 0.11m as a direct value for 'h2')
• r₂ = 0.104 m (derived from sqrt(0.1² + 0.03²) ≈ 0.104 m)
• r₃ = 0.102 m (derived from sqrt(0.1² + 0.02²) ≈ 0.102 m)

System of Equations for Unknown Charges

Two equations are set up based on electric potential (V) at different points, using the formula V = k * Σ(q_i / r_i), where k = 9 × 10⁹ N·m²/C². We are solving for unknown charges Q_B and Q_D.

Equation 1: Potential at -680 V

-680 V = k * (q_A / 0.05 + Q_B / 0.11 + q_C / 0.03 + Q_D / 0.104)

Substituting known values:

• -680 = (9 × 10⁹) * (-6 × 10⁻⁸ / 0.05 + Q_B / 0.11 + 12 × 10⁻⁸ / 0.03 + Q_D / 0.104)
• -680 = -10800 + (8.18 × 10¹⁰ * Q_B) + 36000 + (8.65 × 10¹⁰ * Q_D)
• Simplifying, we get: -25880 = 8.18 × 10¹⁰ * Q_B + 8.65 × 10¹⁰ * Q_D

Equation 2: Potential at -860 V

-860 V = k * (Q_B / 0.02 + Q_D / 0.06 + 1.2 × 10⁻⁸ / 0.11 + -6 × 10⁻⁸ / 0.102)

Substituting known values:

• -860 = (9 × 10⁹) * (Q_B / 0.02 + Q_D / 0.06 + 1.2 × 10⁻⁸ / 0.11 + -6 × 10⁻⁸ / 0.102)
• -860 = (4.5 × 10¹¹ * Q_B) + (1.5 × 10¹¹ * Q_D) + 9818.18 - 5294.11
• Simplifying, we get: -4754.07 = 4.5 × 10¹¹ * Q_B + 1.5 × 10¹¹ * Q_D

Solution for Unknown Charges

Solving the system of these two linear equations yields the values for Q_B and Q_D:

• Q_B ≈ 1.3 × 10⁻⁷ C
• Q_D ≈ -4.22 × 10⁻⁷ C

Electrical Resistance, Power, and Energy

This section covers calculations for electrical resistance, power dissipation, energy consumption, and heat generation in a conductor, considering temperature effects.

Conductor Properties and Resistance

Given a conductor with:

• Length (L): 200 m
• Diameter (D): 0.75 mm
• Initial Temperature (T₀): 20 °C

First, the cross-sectional area (A) is determined. Assuming A = 0.44 mm² (or 0.44 × 10⁻⁶ m²) as given by the calculation `0.75² / 2 = 0.28125` which is then stated as `0.44` (likely a rounded or pre-calculated value for area in mm²).

The initial resistance (R₀) at 20 °C is calculated using resistivity (ρ = 1.72 × 10⁻⁸ Ω·m for copper):

• R₀ = ρ * L / A = (1.72 × 10⁻⁸ Ω·m * 200 m) / (0.44 × 10⁻⁶ m²) ≈ 7.81 Ω

The final resistance (R_f) at 95 °C, considering a temperature coefficient (α = 0.045 / °C), is:

• R_f = R₀ * (1 + α * ΔT) = 7.81 Ω * (1 + 0.045 * (95 - 20)°C)
• R_f = 7.81 Ω * (1 + 0.045 * 75) = 7.81 Ω * (1 + 3.375) = 7.81 Ω * 4.375 ≈ 34.125 Ω

Current, Voltage, and Power Dissipation

With a supply voltage of 250 V and the final resistance (plus an additional 2.5 Ω series resistance), the current (I) is:

• I = V / (R_f + R_additional) = 250 V / (34.125 Ω + 2.5 Ω) = 250 V / 36.625 Ω ≈ 6.8268 A

The voltage drop (V_drop) across the conductor is:

• V_drop = I * R_f = 6.8268 A * 34.125 Ω ≈ 232.93 V

The power (P) dissipated by the conductor is:

• P = V_drop * I = 232.93 V * 6.8268 A ≈ 1590.16 W

Energy Consumption and Heat Generation

Assuming the power is 1.59 kW and operation time is 2 hours, the energy (E) consumed is:

• E = P * t = 1.59 kW * 2 h = 3.18 kWh

The heat (Q) generated in calories, over a period of 100 seconds, is calculated using the conversion factor 0.24 cal/J:

• Q = 0.24 * I² * R_f * t = 0.24 * (6.8268 A)² * 34.125 Ω * 100 s ≈ 38169.65 cal

Total operating time is also noted as 180 hours (6 units * 30 hours/unit).