Answer:
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
