Question

Find the peak current and resonant frequency of the following circuit (as shown in the figure)

• A. $0.2A$and $100Hz$
• B. $2A$and $50Hz$
• C. $2A$and $100Hz$
• D. $0.2A$and $50Hz$

Answer 1

The correct option is D

$0.2A$and $50Hz$

Step 1: Given data.

Inductance is $L=100mH$

Capacitance is $C=100\mu f$

Resistance is $R=120ohm$

Compare the given equation $V=30\sin 100t$ with $V=V_0\sin \omega t$

Therefore, $V_0=30$ and $\omega =100$

Step 2: Calculating $X_L$,$X_c$

We know,

$$\begin{gathered} X_L=\omega L \\ =100\times 100\times {10}^{-3} \\ =10ohm \end{gathered}$$

Where, $X_L$ is the inductive reactance

$\omega$ is the angular frequency

$$\begin{gathered} X_c=\frac{1}{\omega C} \\ =\frac{1}{100\times 100\times {10}^{-6}} \\ =100ohm \end{gathered}$$

Where, $X_c$is the capacitive reactance

$\omega$ is the angular frequency

$C$ is the capacitance

Step 3: Calculating peak current

Peak current in series LCR circuit is $i=\frac{V_0}{Z}$

Where, $i$ is the peak current in the circuit

$Z$ is the impedance and it is given by $Z=\sqrt{\left({\left(X_L-X_c\right)}^2+R^2\right)}$

Where, $X_L$ is the inductive reactance

$X_c$is the capacitive reactance

Therefore

$$\begin{gathered} i=\frac{30}{\sqrt{{(10-100)}^2+{120}^2}} \\ =\frac{30}{150} \\ =0.2Amp \end{gathered}$$

Step 4: Calculating resonant frequency

We know, resonant frequency is given as $f=\frac{1}{2\pi \sqrt{LC}}$

Where, $C$ is the capacitance

$L$is the Inductance

By substituting the values in the above formula, we get

$$\begin{gathered} =\frac{1}{2\pi \sqrt{100\times {10}^{-3}\times 100\times {10}^{-6}}} \\ =\frac{100\sqrt{10}}{2\pi } \\ =\frac{100\pi }{2\pi } \\ =50Hz \end{gathered}$$

Therefore, the correct option is (D).