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.

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