Question:
In an ac circuit containing Resistance R and capacitance C, the current is I. Keeping the ac voltage constant, if the frequency is made \( \frac{1}{3} \), the current is \( \frac{I}{2} \). Then the ratio of initial reactance to the resistance is:
In an ac circuit containing Resistance R and capacitance C, the current is I. Keeping the ac voltage constant, if the frequency is made \( \frac{1}{3} \), the current is \( \frac{I}{2} \). Then the ratio of initial reactance to the resistance is:
Show Hint
Capacitive reactance is inversely proportional to frequency:
\[
X_c=\frac{1}{2\pi fC}.
\]
If the frequency becomes one-third, the capacitive reactance becomes three times its original value.
Updated On: Jun 9, 2026
- \( \left(\frac{3}{5}\right)^{1/2} \)
- \( \left(\frac{2}{5}\right)^{1/2} \)
- \( \left(\frac{1}{5}\right)^{1/2} \)
- \( \left(\frac{4}{5}\right)^{1/2} \)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
For a series RC circuit, the impedance is
\[
Z=\sqrt{R^2+X_c^2},
\]
where
\[
X_c=\frac{1}{2\pi fC}.
\]
The current is given by
\[
I=\frac{V}{Z}.
\]
Since the applied voltage remains constant, any change in current is due to the change in capacitive reactance.
Step 1: Write the expression for the initial current.
Initially, \[ I=\frac{V}{\sqrt{R^2+X_c^2}}. \]
Step 2: Determine the new reactance when frequency is reduced.
Since \[ X_c=\frac{1}{2\pi fC}, \] if the frequency becomes \[ f'=\frac{f}{3}, \] then the new capacitive reactance is \[ X_c'=\frac{1}{2\pi (f/3)C}=3X_c. \] The new current is given to be \[ \frac{I}{2} = \frac{V}{\sqrt{R^2+(3X_c)^2}} = \frac{V}{\sqrt{R^2+9X_c^2}}. \]
Step 3: Form the required equation.
Substituting the value of \(I\), \[ \frac{1}{2}\cdot \frac{V}{\sqrt{R^2+X_c^2}} = \frac{V}{\sqrt{R^2+9X_c^2}}. \] Cancelling \(V\), \[ \sqrt{R^2+9X_c^2} = 2\sqrt{R^2+X_c^2}. \] Squaring both sides, \[ R^2+9X_c^2 = 4(R^2+X_c^2). \] \[ R^2+9X_c^2 = 4R^2+4X_c^2. \] \[ 5X_c^2 = 3R^2. \] Hence, \[ \frac{X_c^2}{R^2} = \frac{3}{5}. \] Therefore, \[ \boxed{\frac{X_c}{R} = \left(\frac{3}{5}\right)^{1/2}} \]
Step 1: Write the expression for the initial current.
Initially, \[ I=\frac{V}{\sqrt{R^2+X_c^2}}. \]
Step 2: Determine the new reactance when frequency is reduced.
Since \[ X_c=\frac{1}{2\pi fC}, \] if the frequency becomes \[ f'=\frac{f}{3}, \] then the new capacitive reactance is \[ X_c'=\frac{1}{2\pi (f/3)C}=3X_c. \] The new current is given to be \[ \frac{I}{2} = \frac{V}{\sqrt{R^2+(3X_c)^2}} = \frac{V}{\sqrt{R^2+9X_c^2}}. \]
Step 3: Form the required equation.
Substituting the value of \(I\), \[ \frac{1}{2}\cdot \frac{V}{\sqrt{R^2+X_c^2}} = \frac{V}{\sqrt{R^2+9X_c^2}}. \] Cancelling \(V\), \[ \sqrt{R^2+9X_c^2} = 2\sqrt{R^2+X_c^2}. \] Squaring both sides, \[ R^2+9X_c^2 = 4(R^2+X_c^2). \] \[ R^2+9X_c^2 = 4R^2+4X_c^2. \] \[ 5X_c^2 = 3R^2. \] Hence, \[ \frac{X_c^2}{R^2} = \frac{3}{5}. \] Therefore, \[ \boxed{\frac{X_c}{R} = \left(\frac{3}{5}\right)^{1/2}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET AC Circuits Questions
- A resistor of resistance \( R \) and an inductor of inductive reactance \( R \) are connected in series to an ac source. A capacitor of capacitive reactance \( 2R \) is then connected in series with \( L \) and \( R \). The ratio of the power factors of LR and LCR circuits is
- An inductor of inductan L, a capacitor of capacitan C and a resistor of resistan R are connected in series to an ac sour. The quality factor of the circuit is
- A resistor of \(50~\Omega\), an inductor and a capacitor are connected in series to an AC source of peak voltage \(200\sqrt{2}~\text{V}\).
When the capacitor alone is removed, current lags voltage by \(37^\circ\).
When the inductor alone is removed, current leads voltage by \(37^\circ\).
The power dissipated in the LCR circuit is: - A \(70 \, \text{mH}\) inductor is connected to a \(220 \, \text{V}, 50 \, \text{Hz}\) AC supply. The RMS value of the current in the circuit is:
- A resistor of 100 \( \Omega \), an inductor of \( \frac{25}{\pi} \) mH, and a capacitor of 0.1 \( \mu \)F are connected in series to an ac source. The impedance of the circuit is minimum for a frequency of:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions