Question

In-circuit$LC$ circuit the inductance$L=40mH$ and $C=100\mu F$. If a voltage $V\left(t\right)=10\sin \left(314t\right)$ is applied to the circuit, the current in the circuit is given as

• A. $10\cos \left(314t\right)$
• B. $0.52\cos \left(314t\right)$
• C. $0.52\sin \left(314t\right)$
• D. $5.2\cos \left(314t\right)$

Answer 1

The correct option is B

$0.52\cos \left(314t\right)$

Solution:

Step 1: Given information and Draw the required diagram

Inductance, $L=40mH$

Capacitance, $C=100\mu F$

Voltage, $V\left(t\right)=10\sin \left(314t\right)$

Step 2: Impedance formula

Use the formula of impedance,

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_c-X_L\right)}^2} \\ Z=\sqrt{{\left(X_c-X_L\right)}^2} \end{gathered}$$

Where, $R=$resistance, $X_c=$Capacitance, and $X_L=$Inductance

Knows, $R=0$, So

$Z=X_C–X_L$

$Z=\left(\frac{1}{\omega C}\right)–\omega L$

Therefore,

$Z=\left[\left(\frac{1}{314}x100x{10}^{-6}\right)\right]–(314x40x{10}^{-3})$

$$\begin{gathered} =31.84–12.56 \\ =19.28ohm \end{gathered}$$

Step 3: Calculate the current in the circuit

$For{X}_C>X_L,currentleadsvoltageby\frac{\pi }{2}$

$$\begin{gathered} i=\frac{V}{Z} \\ i=\frac{10\sin \left(314t+{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{19.28} \end{gathered}$$

$i=0.52\cos \left(314t\right)$

Hence, the correct option is (B).