A series combination of resistor R and capaci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.The initial current is $I = \frac{V}{Z}$.When the frequency cha

Vedclass · Accepted Answer

$\sqrt{0.6}$

Vedclass · Answer

(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.The initial current is $I = \frac{V}{Z}$.When the frequency changes to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3X_c$.The new impedance is $Z' = \sqrt{R^2 + (X_c')^2} = \sqrt{R^2 + (3X_c)^2}$.Given that the new current $I' = \frac{I}{2}$,we have $\frac{V}{Z'} = \frac{1}{2} \frac{V}{Z}$,which implies $Z' = 2Z$.Squaring both sides,$(Z')^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.$5X_c^2 = 3R^2$.$\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.Therefore,the ratio $\frac{X_c}{R} = \sqrt{0.6}$.

Similar Questions

When an $AC$ generator of $120 \, V$ is connected in series with a capacitor and a resistor of $30 \, \Omega$,the circuit carries a current of $1.5 \, A$. The potential difference across the capacitor will be ..... $V$.

In an $LCR$ series $a.c.$ circuit,the voltage across each of the components $L, C$ and $R$ is $60 \,V$. The voltage across the $LC$ combination is

An $a.c.$ circuit contains a resistance of $12 \ \Omega$ and an inductor of inductive reactance $5 \ \Omega$. The phase angle between current and potential difference will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module