Question 54: AC fundamentals

Question

An a.c. source of angular frequency $\omega$ 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 $\omega / 4$, (keeping the voltage unchanged) the current is found to be $I / 3$. The ratio of resistance to reactance at frequency $\omega$ is

Accepted Answer

For a series $RC$ circuit, the current is
$$ I=\frac{V}{Z} $$
where the impedance is
$$ Z=\sqrt{R^2+X_C^2} $$
and capacitive reactance is
$$ X_C=\frac{1}{\omega C} $$
At angular frequency $\omega$, let the reactance be
$$ X=\frac{1}{\omega C} $$
So current is
$$ I=\frac{V}{\sqrt{R^2+X^2}} $$
Now frequency is changed to $\omega/4$.
Then new capacitive reactance becomes
$$ X_C'=\frac{1}{(\omega/4)C}=\frac{4}{\omega C}=4X $$
So new current is
$$ I'=\frac{V}{\sqrt{R^2+(4X)^2}}=\frac{V}{\sqrt{R^2+16X^2}} $$
Given that
$$ I'=\frac{I}{3} $$
Substitute expressions for $I'$ and $I$:
$$ \frac{V}{\sqrt{R^2+16X^2}}=\frac{1}{3}\cdot \frac{V}{\sqrt{R^2+X^2}} $$
Cancel $V$:
$$ \frac{1}{\sqrt{R^2+16X^2}}=\frac{1}{3\sqrt{R^2+X^2}} $$
So,
$$ 3\sqrt{R^2+X^2}=\sqrt{R^2+16X^2} $$
Squaring both sides:
$$ 9(R^2+X^2)=R^2+16X^2 $$
$$ 9R^2+9X^2=R^2+16X^2 $$
$$ 8R^2=7X^2 $$
Therefore,
$$ \frac{R^2}{X^2}=\frac{7}{8} $$
Hence,
$$ \frac{R}{X}=\sqrt{\frac{7}{8}} $$
So the required ratio of resistance to reactance at frequency $\omega$ is
$$ \boxed{\sqrt{\frac{7}{8}}} $$
Therefore, the correct option is:
$$ \boxed{	ext{Option C}} $$

Answer

$\sqrt{\frac{6}{7}}$

Answer

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

Answer

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

Answer

$$ 	ext { } \sqrt{\frac{3}{4}} $$

AC fundamentals

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