A series R-C circuit is connected to an AC vo | vedclass

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Question

(A) In a series $R-C$ circuit,the impedance is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.When a dielectric of constant $K = 4

Vedclass · Accepted Answer

$(B, C)$

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(A) In a series $R-C$ circuit,the impedance is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.When a dielectric of constant $K = 4$ is inserted,the new capacitance becomes $C' = KC = 4C$.Consequently,the new capacitive reactance becomes $X_C' = \frac{1}{\omega (4C)} = \frac{X_C}{4}$.Since $X_C' The current in the circuit is $I = \frac{V}{Z}$. Since $Z'  I_R^A$ (Option $B$ is true).The voltage across the capacitor is $V_C = I X_C = \frac{V}{\sqrt{R^2 + X_C^2}} X_C = \frac{V}{\sqrt{(R/X_C)^2 + 1}}$.As $X_C$ decreases,the term $(R/X_C)^2$ increases,which makes the denominator $\sqrt{(R/X_C)^2 + 1}$ larger.Therefore,$V_C$ decreases when $X_C$ decreases. Thus,$V_C^A > V_C^B$ (Option $C$ is true).Hence,the correct statements are $(B)$ and $(C)$.

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