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Lv 6
? asked in Science & MathematicsEngineering · 9 years ago

electrical engineering ac analysis?

A series circuit consists of a coil, a resistor and a capacitor connected across a 250 V, 60 Hz power supply. The voltage across the coil is 150 V, that across the resistor is 100V and that across the capacitor is also 100 V. THe resultant voltage across the coil and the resistor is 200 V. If the circuit draws 15 A from the source, determine (a) resistance and inductance of the coil (b) the resistance and capacitance of the capacitor.

Update:

to Douglas, your final answers are wrong. sorry...

2 Answers

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  • 9 years ago
    Favorite Answer

    OK.... I blew it on the first try. Here is a second attempt.

    The magnitude of the voltage across the inductor (V_l) squared divided by the magnitude of the current (I) squared equals the sum of the squares of the resistive component within the coil (R_l) and the inductive impedance of the coil (2π(f)(L)):

    (V_l)²/I² = R_l² + {2π(f)(L)}²

    The magnitude of the voltage of across the inductor and resistor (V_l-R) squared divided by the magnitude for the current (I) squared equals the sum of the squares of the resistive components and the inductive components

    V_l-R²/I² = (R + R_l)² + {2π(f)(L)}²

    Subtracting the first equation from the second will eliminate the inductive component:

    (V_l-R² - V_l²)/I² = (R + R_l)² - R_l²

    Squaring the first term on the right:

    (V_l-R² - V_l²)/I² = R² + 2R(R_l) + R_l² - R_l²

    Combining like terms:

    (V_l-R² - V_l²)/I² = R² + 2R(R_l)

    Substituting V_r/I for R:

    (V_l-R² - V_l²)/I² = V_r²/I² + 2(V_r/I)(R_l)

    Subtract V_r²/I² from both sides:

    (V_l-R² - V_l²)/I² - V_r²/I² = 2(V_r/I)(R_l)

    Regrouping:

    (V_l-R² - V_l² - V_r²)/I² = 2(V_r/I)(R_l)

    Multiply both sides by I:

    (V_l-R² - V_l² - V_r²)/I = 2(V_r)(R_l)

    Divide both side 2(V_r):

    R_l = (V_l-R² - V_l² - V_r²)/{2I(V_r)}

    Substituting values for the right-hand side:

    R_l = (200² - 150² - 100²)/{2(15)(100)}

    R_l = (40000 - 22500 - 10000)/(3000) = 7500/3000 = 2.5 Ω

    The resistive component of the coil (R_l) is 2.5 Ω

    We may find the inductive component of the coil by returning to the equation:

    (V_l)²/I² = R_l² + {2π(f)(L)}²

    Subtracting R_l² from both sides:

    {2π(f)(L)}² = (V_l)²/I² - R_l²

    Take the square root of both side:

    2π(f)(L) = sqrt{(V_l)²/I² - R_l²}

    L = sqrt{(V_l)²/I² - R_l²}/2π(f)

    Substituting values for the right-hand side:

    L = sqrt{(150)²/15² - 2.5²}/2π(60)

    L = sqrt(100 - 6.25)/120π

    L = sqrt(93.75)/120π

    L ≈ 2.568 x 10^-2 H

    Let's find the phase angle for the impedance of resistor and the inductor:

    Θ = tan^-1{sqrt(93.75)/(100/15 + 2.5)} = 46.567°

    This means the phase angle for the capacitor and parasitic resistance must have the same phase but opposite in sign:

    Θ = -46.567°

    The resistive component (R_c) is the magnitude of the voltage (V_c) divided by the magnitude of the current multiplied by (I) multiplied by the cosine of the phase angle:

    R_c = (V_c/I)cos(Θ)

    Substituting values for the right-hand side:

    R_c = (100/15)cos(-46.567) ≈ 4.58 Ω

    The capacitive component (-1/{2π(f)C}) is equal the magnitude of the voltage (V_c) divided by the the magnitude of the current multiplied by the sine of the phase angle:

    -1/{2π(f)C} = (V_c/I)sin(Θ)

    2π(f)C = -I/{(V_c)sin(Θ)}

    C = -I/{2π(f)(V_c)sin(Θ)}

    Substituting values for the right-hand side:

    C = -15/{120π(100)sin(-46.567)} ≈ 5.479 x 10^-4 F

    Check:

    The impedance of the inductor is:

    sqrt{2.5² + (120π(2.568 x 10^-2)²} ≈ sqrt(100) = 10

    The impedance of the capacitor is:

    sqrt{4.58² + (1/{120π(5.479 x 10^-4)})²} ≈ 20/3

    When multiplied by 15 A, both of these values give the respective voltage. Therefore, this checks.

  • ?
    Lv 4
    4 years ago

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