Question 183: AC fundamentals

Question

A series LCR circuit with $R=20 \Omega, L=1.6 \mathrm{H}$ and $C=40 \mu \mathrm{~F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is $\_\_\_\_$ $\Omega$.

Accepted Answer

In a series $L C R$ circuit, resonance occurs when the inductive reactance $\left(X_Light)$ is equal to the capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}ight)$.
$$ X_L=X_C $$
$$\Rightarrow $$ $$ \omega_{\mathrm{r}} \mathrm{~L}=\frac{1}{\omega_{\mathrm{r}} \mathrm{C}} $$
Where $\omega_{\mathrm{r}}$ is the resonant angular frequency.
$$ \omega_{\mathrm{r}}^2=\frac{1}{\mathrm{LC}} $$
$$\Rightarrow $$ $$ \omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{LC}}} $$
The inductance of inductor is given as $\mathrm{L}=1.6 \mathrm{H}$ and the capacitance of capacitor is $\mathrm{C}=40 \mu \mathrm{~F}=40 	imes 10^{-6} \mathrm{~F}$
Putting the values,
$$ \omega_{\mathrm{r}}=\frac{1}{\sqrt{1.6 	imes 40 	imes 10^{-6}}} $$
$$ \begin{gathered} \omega_{\mathrm{r}}=\frac{1}{\sqrt{64 	imes 10^{-6}}} \ \omega_{\mathrm{r}}=\frac{1}{8 	imes 10^{-3}}=\frac{1000}{8}=125 \mathrm{rad} / \mathrm{s} \end{gathered} $$
Using the formula for inductive reactance
$$ \begin{gathered} \mathrm{X}_{\mathrm{L}}=\omega_{\mathrm{r}} \cdot \mathrm{~L} \ \mathrm{X}_{\mathrm{L}}=125 	imes 1.6 \ \mathrm{X}_{\mathrm{L}}=125 	imes \frac{16}{10}=200 \Omega \end{gathered} $$
Therefore, the inductive reactance at the resonant frequency is $200 \Omega$.

AC fundamentals

A series LCR circuit with R=20 \Omega, L=1.6 \mathrm{H} and C=40 \mu \mathrm{~F} is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is \_\_\_\_ \Omega.

Integer / Numeric Answer Type