Answer:
Xc = (0.467 - 0.427j)R
Step-by-step explanation:
Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is
Z = √[R² + (XL - XC)²]
Since the inductive reactance XL equals the resistance R, we have that
Z = √[R² + (XL - XC)²]
Z = √[R² + (R - XC)²]
Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]
Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus
Z' = √[R² + (R - XC')²]
Z' = √[R² + (R - 2XC)²]
The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]
Since the current doubles, I' = 2I.
V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]
1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]
√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]
squaring both sides, we have
[R² + (R - XC)²] = 4[R² + (R - 2XC)²]
expanding the brackets, we have
[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]
[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]
2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²
collecting like terms, we have
16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²
14RXC - 15XC² = 6R²
15XC² - 14RXC + 6R² = 0
Using the quadratic formula to find XC, we have
Since it is capacitive, we take the negative part.
So, Xc = (0.467 - 0.427j)R