An alternating voltage of 240V, 50Hz is connected across an impedance of (60-1100)Ω. Determine (a) the resistance (b) the capacitance (c) the magnitude of the impedance and its phase angle and (d) the current flowing.

Question

Asked on July 2, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-02T09:12:39.339Z

Let's solve the questions step-by-step. Q1: Determine the resistance, capacitance, magnitude of the impedance, phase angle, and current flowing. Given: Voltage \( V = 240 \, V \) Frequency \( f = 50 \, Hz \) Impedance \( Z = 60 - j100 \, \) a) Resistance \( R \): The real part of the impedance is the resistance. \[ R = 60 \, \] b) Capacitance \( C \): The imaginary part of the impedance is related to the capacitive reactance \( X_C \). \[ X_C = -100 \, \] \[ X_C = (1)/(2 f C) \] \[ C = (1)/(2 f |X_C|) = (1)/(2 × 50 × 100) \] \[ C = (1)/(10000) \, F \] c) Magnitude of the impedance \( |Z| \): \[ |Z| = sqrt(R^2 + X_C^2) = sqrt(60^2 + 100^2) \] \[ |Z| = sqrt(3600 + 10000) = sqrt(13600) \] \[ |Z| = 116.62 \, \] d) Phase angle \( \): \[ = ^-1((X_C)/(R)) = ^-1((-100)/(60)) \] \[ = ^-1(-(5)/(3)) ≈ -59.04^ \] e) Current \( I \): \[ I = (V)/(|Z|) = (240)/(116.62) \] \[ I ≈ 2.06 \, A \] Final Answers: Resistance: \( 60 \, \) Capacitance: \( (1)/(10000) \, F \) Magnitude of Impedance: \( 116.62 \, \) Phase Angle: \( -59.04^ \) Current: \( 2.06 \, A \) Q2: Lagrange Interpolation to find \( f(1.25) \). Given data points: \((x_0, y_0) = (0, 1)\), \((x_1, y_1) = (1, 2.25)\), \((x_2, y_2) = (2, 3.75)\), \((x_3, y_3) = (3, 4.25)\). The Lagrange interpolation formula is: P(x) = _j=0^n y_j L_j(x) where L_j(x) = _k=0, k ≠ j^n (x - x_k)/(x_j - x_k) Calculate \( L_j(1.25) \): 1. \( L_0(1.25) = ((1.25-1)(1.25-2)(1.25-3))/((0-1)(0-2)(0-3)) = ((-0.75)(-0.75)(-1.75))/((-1)(-2)(-3)) = (-0.984375)/(6) = -0.1640625 \) 2. \( L_1(1.25) = ((1.25-0)(1.25-2)(1.25-3))/((1-0)(1-2)(1-3)) = ((1.25)(-0.75)(-1.75))/((1)(-1)(-2)) = (1.640625)/(2) = 0.8203125 \) 3. \( L_2(1.25) = ((1.25-0)(1.25-1)(1.25-3))/((2-0)(2-1)(2-3)) = ((1.25)(0.25)(-1.75))/((2)(1)(-1)) = (-0.546875)/(-2) = 0.2734375 \) 4. \( L_3(1.25) = ((1.25-0)(1.25-1)(1.25-2))/((3-0)(3-1)(3-2)) = ((1.25)(0.25)(-0.75))/((3)(2)(1)) = (-0.234375)/(6) = -0.0390625 \) Calculate \( P(1.25) \): P(1.25) = 1(-0.1640625) + 2.25(0.8203125) + 3.75(0.2734375) + 4.25(-0.0390625) P(1.25) = -0.1640625 + 1.845703125 + 1.025390625 - 0.166015625 P(1.25) = 2.541015625 Final Answer: 2.541 Q3: Use Gauss elimination method to solve the system of equations. The system of equations is: 1. \( x_1 - x_2 + x_3 = 0 \) 2. \( -x_1 + x_2 - x_3 = 0 \) 3. \( 10x_1 + 2x_2 - x_3 = 90 \) 4. \( 20x_2 + 10x_3 = 80 \) Step 1: Write the augmented matrix. \[ 1 & -1 & 1 & | & 0 \\ -1 & 1 & -1 & | & 0 \\ 10 & 2 & -1 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Step 2: Perform row operations to get an upper triangular matrix. Add Row 1 to Row 2: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 10 & 2 & -1 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Replace Row 3 with Row 3 - 10*Row 1: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 12 & -11 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Replace Row 4 with Row 4 - \((5)/(3)\)*Row 3: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 12 & -11 & | & 90 \\ 0 & 0 & (130)/(3) & | & (200)/(3) \\ \] Step 3: Back substitution. From Row 4: \[ x_3 = (200)/(130) = (20)/(13) \] From Row 3: \[ 12x_2 - 11x_3 = 90 \] \[ 12x_2 - 11((20)/(13)) = 90 \] \[ 12x_2 = 90 + (220)/(13) \] \[ x_2 = (90 × 13 + 220)/(12 × 13) \] \[ x_2 = (1430)/(156) = (715)/(78) \] From Row 1: \[ x_1 - x_2 + x_3 = 0 \] \[ x_1 = x_2 - x_3 \] \[ x_1 = (715)/(78) - (20)/(13) \] \[ x_1 = (715)/(78) - (120)/(78) \] \[ x_1 = (595)/(78) \] Final Answers: \( x_1 \): \( (595)/(78) \) \( x_2 \): \( (715)/(78) \) \( x_3 \): \( (20)/(13) \) What's next? 📸

Accepted Answer · 2026-07-02T09:12:39.339Z

Let's solve the questions step-by-step. Q1: Determine the resistance, capacitance, magnitude of the impedance, phase angle, and current flowing. Given: Voltage \( V = 240 \, V \) Frequency \( f = 50 \, Hz \) Impedance \( Z = 60 - j100 \, \) a) Resistance \( R \): The real part of the impedance is the resistance. \[ R = 60 \, \] b) Capacitance \( C \): The imaginary part of the impedance is related to the capacitive reactance \( X_C \). \[ X_C = -100 \, \] \[ X_C = (1)/(2 f C) \] \[ C = (1)/(2 f |X_C|) = (1)/(2 × 50 × 100) \] \[ C = (1)/(10000) \, F \] c) Magnitude of the impedance \( |Z| \): \[ |Z| = sqrt(R^2 + X_C^2) = sqrt(60^2 + 100^2) \] \[ |Z| = sqrt(3600 + 10000) = sqrt(13600) \] \[ |Z| = 116.62 \, \] d) Phase angle \( \): \[ = ^-1((X_C)/(R)) = ^-1((-100)/(60)) \] \[ = ^-1(-(5)/(3)) ≈ -59.04^ \] e) Current \( I \): \[ I = (V)/(|Z|) = (240)/(116.62) \] \[ I ≈ 2.06 \, A \] Final Answers: Resistance: \( 60 \, \) Capacitance: \( (1)/(10000) \, F \) Magnitude of Impedance: \( 116.62 \, \) Phase Angle: \( -59.04^ \) Current: \( 2.06 \, A \) Q2: Lagrange Interpolation to find \( f(1.25) \). Given data points: \((x_0, y_0) = (0, 1)\), \((x_1, y_1) = (1, 2.25)\), \((x_2, y_2) = (2, 3.75)\), \((x_3, y_3) = (3, 4.25)\). The Lagrange interpolation formula is: P(x) = _j=0^n y_j L_j(x) where L_j(x) = _k=0, k ≠ j^n (x - x_k)/(x_j - x_k) Calculate \( L_j(1.25) \): 1. \( L_0(1.25) = ((1.25-1)(1.25-2)(1.25-3))/((0-1)(0-2)(0-3)) = ((-0.75)(-0.75)(-1.75))/((-1)(-2)(-3)) = (-0.984375)/(6) = -0.1640625 \) 2. \( L_1(1.25) = ((1.25-0)(1.25-2)(1.25-3))/((1-0)(1-2)(1-3)) = ((1.25)(-0.75)(-1.75))/((1)(-1)(-2)) = (1.640625)/(2) = 0.8203125 \) 3. \( L_2(1.25) = ((1.25-0)(1.25-1)(1.25-3))/((2-0)(2-1)(2-3)) = ((1.25)(0.25)(-1.75))/((2)(1)(-1)) = (-0.546875)/(-2) = 0.2734375 \) 4. \( L_3(1.25) = ((1.25-0)(1.25-1)(1.25-2))/((3-0)(3-1)(3-2)) = ((1.25)(0.25)(-0.75))/((3)(2)(1)) = (-0.234375)/(6) = -0.0390625 \) Calculate \( P(1.25) \): P(1.25) = 1(-0.1640625) + 2.25(0.8203125) + 3.75(0.2734375) + 4.25(-0.0390625) P(1.25) = -0.1640625 + 1.845703125 + 1.025390625 - 0.166015625 P(1.25) = 2.541015625 Final Answer: 2.541 Q3: Use Gauss elimination method to solve the system of equations. The system of equations is: 1. \( x_1 - x_2 + x_3 = 0 \) 2. \( -x_1 + x_2 - x_3 = 0 \) 3. \( 10x_1 + 2x_2 - x_3 = 90 \) 4. \( 20x_2 + 10x_3 = 80 \) Step 1: Write the augmented matrix. \[ 1 & -1 & 1 & | & 0 \\ -1 & 1 & -1 & | & 0 \\ 10 & 2 & -1 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Step 2: Perform row operations to get an upper triangular matrix. Add Row 1 to Row 2: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 10 & 2 & -1 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Replace Row 3 with Row 3 - 10*Row 1: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 12 & -11 & | & 90 \\ 0 & 20 & 10 & | & 80 \\ \] Replace Row 4 with Row 4 - \((5)/(3)\)*Row 3: \[ 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 12 & -11 & | & 90 \\ 0 & 0 & (130)/(3) & | & (200)/(3) \\ \] Step 3: Back substitution. From Row 4: \[ x_3 = (200)/(130) = (20)/(13) \] From Row 3: \[ 12x_2 - 11x_3 = 90 \] \[ 12x_2 - 11((20)/(13)) = 90 \] \[ 12x_2 = 90 + (220)/(13) \] \[ x_2 = (90 × 13 + 220)/(12 × 13) \] \[ x_2 = (1430)/(156) = (715)/(78) \] From Row 1: \[ x_1 - x_2 + x_3 = 0 \] \[ x_1 = x_2 - x_3 \] \[ x_1 = (715)/(78) - (20)/(13) \] \[ x_1 = (715)/(78) - (120)/(78) \] \[ x_1 = (595)/(78) \] Final Answers: \( x_1 \): \( (595)/(78) \) \( x_2 \): \( (715)/(78) \) \( x_3 \): \( (20)/(13) \) What's next? 📸

An alternating voltage of 240V, 50Hz is connected across an impedance of (60-1100)Ω. Determine (a) the resistance (b) the capacitance (c) the magnitude of the impedance and its phase angle and (d) the current flowing.

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An alternating voltage of 240V, 50Hz is connected across an impedance of (60-1100)Ω. Determine (a) the resistance (b) the capacitance (c) the magnitude of the impedance and its phase angle and (d) the current flowing.

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