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

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R =

where

R = resistance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

where

x = separation between the parallel plates

Thus

C ∝

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,

Also,

Z ∝ I

Therefore,

Solving the above eqn: