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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?

Respuesta :

Answer:

Case i) if [tex]\omega L > \frac{1}{\omega c}[/tex]

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

Case ii) if [tex]\omega L < \frac{1}{\omega c}[/tex]

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

[tex]z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}[/tex]

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

[tex]c_{eq} = c_1 + c_2[/tex]

so it is

[tex]c_{eq} = 2c[/tex]

now we have

[tex]z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}[/tex]

Case i) if [tex]\omega L > \frac{1}{\omega c}[/tex]

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

Case ii) if [tex]\omega L < \frac{1}{\omega c}[/tex]

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

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