Q.44. An a.c. source of angular frequency is connected across a resistor R and a capacitor C in ser

Q: What is the correct answer for this exam Physics Q44?

The correct answer is C. Sol. For an RC circuit, the impedance is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{\omega C}$. The current is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$. At frequency $\omega$: $I = \frac{V}{\sqrt{R^2 + X_C^2}}$. At frequency $\omega' = \omega/4$: $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/4) C} = \frac{4}{\omega C} = 4X_C$. The new current is $I' = \frac{V}{\sqrt{R^2 + (4

Q: Q.44. An a.c. source of angular frequency is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency of the source is changed to , (keeping the…

The correct answer is C. Sol. For an RC circuit, the impedance is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{\omega C}$. The current is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$. At frequency $\omega$: $I = \frac{V}{\sqrt{R^2 + X_C^2}}$. At frequency $\omega' = \omega/4$: $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/4) C} = \frac{4}{\omega C} = 4X_C$. The new current is $I' = \frac{V}{\sqrt{R^2 + (4

Q: What are the options for this this exam Physics question?

A. $\sqrt{\frac{3}{5}}$ · B. $\sqrt{\frac{2}{5}}$ · C. $\sqrt{\frac{7}{8}}$ · D. $\sqrt{\frac{1}{4}}$

Question

Q.44. An a.c. source of angular frequency  is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency of the source is changed to , (keeping the voltage unchanged) the current is found to be I/3. The ratio of resistance to reactance at frequency  is

FirstInTest · Accepted Answer

Sol. For an RC circuit, the impedance is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{\omega C}$. The current is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$. At frequency $\omega$: $I = \frac{V}{\sqrt{R^2 + X_C^2}}$. At frequency $\omega' = \omega/4$: $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/4) C} = \frac{4}{\omega C} = 4X_C$. The new current is $I' = \frac{V}{\sqrt{R^2 + (4X_C)^2}}$. Given $I' = I/3$. So, $\frac{V}{\sqrt{R^2 + (4X_C)^2}} = \frac{1}{3} \frac{V}{\sqrt{R^2 + X_C^2}}$. Squaring both sides: $\frac{1}{R^2 + 16X_C^2} = \frac{1}{9(R^2 + X_C^2)}$. $9(R^2 + X_C^2) = R^2 + 16X_C^2$. $9R^2 + 9X_C^2 = R^2 + 16X_C^2$. $8R^2 = 7X_C^2$. $\frac{R^2}{X_C^2} = \frac{7}{8}$. $\frac{R}{X_C} = \sqrt{\frac{7}{8}}$.

Answer

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

Answer

$\sqrt{\frac{2}{5}}$

Answer

$\sqrt{\frac{7}{8}}$

Answer

$\sqrt{\frac{1}{4}}$

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