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1 November, 07:47

A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-fourth its original value, the current in the circuit quadruples. Determine the initial capacitive reactance in terms of the resistance R.

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  1. 1 November, 08:13
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    Xc₁ = 5R

    The initial capacitance Xc₁ is 5 times the resistance R

    Explanation:

    Let Z₁ is the initial impedance before the separation of parallel plate capacitor is reduced reduced to one-fourth its original value.

    Let Z₂ it the final impedance after the separation of parallel plate capacitor is reduced reduced to one-fourth its original value and current is quadrupled.

    Since the current is quadrupled after the separation, the impedance Z₂ is reduced to one-fourth as compared to impedance Z₁

    Z₂/Z₁ = 1/4

    Since impedance is given by

    Z = √R² + (XL-Xc) ²

    So the above relation becomes

    √R² + (XL-Xc₂) ²/√R² + (XL-Xc₁) ² = 1/4

    Also it is given that R is equal to the inductive reactance XL

    √R² + (R-Xc₂) ²/√R² + (R-Xc₁) ² = 1/4

    As we know Xc = 1/2πfC

    The capacitive reactance has inverse relation with capacitance and the capacitance has also inverse relation with separation of plates, therefore, the capacitive reactance Xc₂ would be one-fourth of Xc₁

    Xc₂ = (1/4) Xc₁

    So the above equation becomes

    √R² + (R-1/4Xc₁) ²/√R² + (R-Xc₁) ² = 1/4

    Squaring both sides

    R² + (R-1/4Xc₁) ²/R² + (R-Xc₁) ² = (1/4) ²

    Simplifying the equation,

    R² + (R² - 2*R*0.25Xc₁+0.0625Xc₁²) / R² + (R² - 2*R*Xc₁+Xc₁²) = 0.0625

    R² + R² - 0.5RXc₁ + 0.0625Xc₁² / R² + R² - 2RXc₁ + Xc₁² = 0.0625

    2R² - 0.5RXc₁ + 0.0625Xc₁² / 2R² - 2RXc₁ + Xc₁² = 0.0625

    2R² - 0.5RXc₁ + 0.0625Xc₁² = 0.0625 (2R² - 2RXc₁ + Xc₁²)

    2R² - 0.5RXc₁ + 0.0625Xc₁² = 0.125R² - 0.125RXc₁ + 0.0625Xc₁²

    2R² - 0.125R² + 0.5RXc₁ + 0.125RXc₁ + 0.0625Xc₁² - 0.0625Xc₁² = 0

    1.875R² - 0.375RXc₁ = 0

    R (1.875R - 0.375Xc₁) = 0

    1.875R - 0.375Xc₁ = 0

    - 0.375Xc₁ = - 1.875R

    Xc₁ = (1.875/0.375) R

    Xc₁ = 5R

    Therefore, the initial capacitance Xc₁ is 5 times the resistance R
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