A series combination of resistor (R) and capa | vedclass

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(A) The impedance of the circuit is $Z = \sqrt{R^2 + X_C^2}$.When the angular frequency is changed to $\omega' = \omega / 3$,the capacitive reactance

Vedclass · Accepted Answer

$\sqrt{0.6}$

Vedclass · Answer

(A) The impedance of the circuit is $Z = \sqrt{R^2 + X_C^2}$.When the angular frequency is changed to $\omega' = \omega / 3$,the capacitive reactance becomes $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3 X_C$.The new impedance is $Z' = \sqrt{R^2 + (3X_C)^2} = \sqrt{R^2 + 9X_C^2}$.Given that the current becomes half,$I' = I/2$,which implies $Z' = 2Z$.Substituting the expressions,we get $2 \sqrt{R^2 + X_C^2} = \sqrt{R^2 + 9X_C^2}$.Squaring both sides,$4(R^2 + X_C^2) = R^2 + 9X_C^2$.$4R^2 + 4X_C^2 = R^2 + 9X_C^2$.$3R^2 = 5X_C^2$.Therefore,$\frac{X_C^2}{R^2} = \frac{3}{5} = 0.6$.Thus,the ratio $\frac{X_C}{R} = \sqrt{0.6}$.

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