Question

In a RLC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit?

Answer 1

Answer:

Case i) if $\omega L > \frac{1}{\omega c}$

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if $\omega L < \frac{1}{\omega c}$

So initially if the circuit is capacitive in nature then its net impedance will increase after this

Explanation:

As we know that the impedance of the circuit is given as

$z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}$

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

$c_{eq} = c_1 + c_2$

so it is

$c_{eq} = 2c$

now we have

$z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}$

Case i) if $\omega L > \frac{1}{\omega c}$

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if $\omega L < \frac{1}{\omega c}$

So initially if the circuit is capacitive in nature then its net impedance will increase after this