Use resonance condition for a series $LCR$ circuit

At resonance, $X_L = X_C$ and the impedance becomes minimum: $Z = R$

Hence the peak current is $I_0 = \frac{V_0}{R}$

Read values from the figure

From the given circuit figure:

• Peak voltage, $V_0 = 10\,\text{V}$
• Resistance, $R = 50\,\Omega$

Therefore, $I_0 = \frac{10}{50} = 0.2\,\text{A}$

Calculate resonant frequency

Resonant frequency is $f_0 = \frac{1}{2\pi\sqrt{LC}}$

From the figure:

• $L = 0.1\,\text{H}$
• $C = 100\,\mu\text{F} = 100 \times 10^{-6}\,\text{F}$

So, $f_0 = \frac{1}{2\pi\sqrt{(0.1)(100\times 10^{-6})}}$

$= \frac{1}{2\pi\sqrt{10^{-5}}}$

$= \frac{1}{2\pi \times 10^{-2.5}}$

Numerically, $\sqrt{10^{-5}} \approx 3.162\times 10^{-3}$

Thus, $f_0 \approx \frac{1}{2\pi \times 3.162\times 10^{-3}} \approx 50\,\text{Hz}$

Match with options

• Peak current $= 0.2\,\text{A}$
• Resonant frequency $= 50\,\text{Hz}$

Therefore, the correct option is: $\boxed{\text{D: } 0.2\,\text{A and } 50\,\text{Hz}}$

Comparison with stored answer

Stored correct answer: D

My derived answer is also D.