An AC source of angular frequency omega is co | vedclass

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

Question

(A) At angular frequency $\omega$,the current $I$ in the $RC$ series circuit is given by:$I = \frac{V}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{R^2 + (\fr

Vedclass · Accepted Answer

$\sqrt{\frac{3}{5}}$

Vedclass · Answer

(A) At angular frequency $\omega$,the current $I$ in the $RC$ series circuit is given by:$I = \frac{V}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{R^2 + (\frac{1}{\omega C})^2}}$  ..........$(i)$When the frequency is changed to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/3)C} = \frac{3}{\omega C} = 3X_C$.Given that the new current $I' = \frac{I}{2}$,we have:$\frac{I}{2} = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$  ..........$(ii)$Dividing equation $(i)$ by $(ii)$:$2 = \frac{\sqrt{R^2 + 9X_C^2}}{\sqrt{R^2 + X_C^2}}$Squaring both sides:$4 = \frac{R^2 + 9X_C^2}{R^2 + X_C^2}$$4R^2 + 4X_C^2 = R^2 + 9X_C^2$$3R^2 = 5X_C^2$$\frac{X_C^2}{R^2} = \frac{3}{5}$$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$

Similar Questions

If $E = 200\,V$,$R = 25\,\Omega$,$L = 2\,H$,and $C = 2\,\mu F$ and the frequency is variable,then the current at $f = 0$ and $f = \infty$ will be respectively:

In the given circuit,the potential difference across the resistance is $54 \, V$ and the power consumed by it is $16 \, W$. If the $AC$ frequency is $60 \, Hz$,find the value of $L$ in $H$.

$A$ student connects a long air-cored coil of manganin wire to a $100\,V$ $D.C.$ supply and records a current of $25\,A$. When the same coil is connected across a $100\,V$,$50\,Hz$ $A.C.$ supply,the current reduces to $20\,A$. The reactance of the coil is....$\Omega $

In an $ac$ circuit,the potential difference across an inductance and resistance joined in series are respectively $16\, V$ and $20\, V$. The total potential difference across the circuit is.......$V$

Vedclass Test Series

Exam Paper Generator

Online Exam Module