Respuesta :
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = [tex]X_{L} = j\omega L = 2\pi fL[/tex]
where
R = resistance
[tex]X_{L} = Inductive Reactance[/tex]
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
[tex]C = \frac{\epsilon_{o}A}{x}[/tex]
where
x = separation between the parallel plates
Thus
C ∝ [tex]\frac{1}{x}[/tex]
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
[tex]Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}[/tex]
Also,
Z ∝ I
Therefore,
[tex]\frac{Z}{I} = \frac{Z'}{I'}[/tex]
[tex]\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}[/tex]
[tex]{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})[/tex]
[tex]{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})[/tex]
Solving the above eqn:
[tex]X_{C} = 4R[/tex]
