Question

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance R is equal to the inductive reactance. If the plate separation of the parallel-plate capacitor is reduced to one-third its original value, the current in the circuit triples. Find the initial capacitive reactance in terms of R

Answer 1

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R =
`X_(L) = j\omega L = 2\pi fL`

where

R = resistance

`X_(L) = Inductive Reactance`

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

`C = (\epsilon_(o)A)/(x)`

where

x = separation between the parallel plates

Thus

C ∝
`(1)/(x)`

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,

`Z = \sqrt{R^(2) + (X_(L) - X_(C))^(2)}`

Also,

Z ∝ I

Therefore,

`(Z)/(I) = (Z')/(I')`

`\frac{\sqrt{R^(2) + (R - X_(C))^(2)}}{3I} = \frac{\sqrt{R^(2) + (R - (X_(C))/(3))^(2)}}{I}`

`{R^(2) + (R - X_(C))^(2)} = 9({R^(2) + (R - (X_(C))/(3))^(2)})`

`{R^(2) + R^(2) + X_(C)^(2) - 2RX_(C) = 9({R^(2) + R^(2) + (X_(C)^(2))/(9) - 2RX_(C))`

Solving the above eqn:

`X_(C) = 4R`