Question 186: AC fundamentals

Question

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V=5 \sin (100 t)$ The values of $L$ and $R$ are shown in the figure. The capacitance of the capacitor ( C ) used is $\_\_\_\_$ $\mu \mathrm{F}$.

Accepted Answer

In a series LCR circuit, the current reaches its maximum value when the circuit is at resonance.
At resonance the inductive reactance $\left(X_Light)$ equals the capacitive reactance $\left(X_Cight)$.
The total impedance $(\mathrm{Z})$ is $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight)^2}=\mathrm{R}$, purely resistive and minimum.
So, the peak current is given by $\mathrm{I}_0=\frac{\mathrm{V}_0}{\mathrm{Z}}=\frac{\mathrm{V}_0}{\mathrm{R}}$.
From the given equation of a.c. voltage is $\mathrm{V}=5 \sin (100 \mathrm{t})$. So, the peak voltage is $\mathrm{V}_0=5 \mathrm{~V}$
The resistance is $\mathrm{R}=100 \Omega=\mathrm{Z}$.
So, the calculated value of peak current is
$$ \mathrm{I}_0=\frac{5}{100}=0.05 \mathrm{~A}=50 \mathrm{~mA} . $$
This shows that the circuit is in resonance state.
From the voltage equation $\mathrm{V}=5  \sin (100 \mathrm{t})$,
$$ \omega=100 \mathrm{rad} / \mathrm{s} $$
Using the resonance condition $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ :
$\omega L=\frac{1}{\omega C}$
$$\Rightarrow $$ $$ C=\frac{1}{\omega^2 L} $$
Putting the values, we get
$$ C=\frac{1}{(100)^2 	imes 2}=\frac{1}{20000} F=\frac{10^6}{20000} \mu F=50 \mu F $$
Therefore, the capacitance of the capacitor is $50 \mu \mathrm{~F}$. Hence, the correct answer is 50.

AC fundamentals

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for V=5 \sin (100 t) The values of L and R are shown in the figure. The capacitance of the capacitor ( C ) used is \_\_\_\_ \mu \mathrm{F}.

Integer / Numeric Answer Type