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.

Question

Physics

Raelynn Vargas

11 January, 16:49

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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Answers (1)

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Samir Gomez · Answer 1

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