1. Initial Data Extraction

Based on the provided equations derived from the graph:

• Current Equation: $I = 400 \text{ mA} \sin(\omega t) = 0.4 \text{ A} \sin(\omega t)$.
  Peak Current ($I_m$) = $0.4 \text{ A}$. Phase $\phi_i = 0$.
• Voltage Equation: $V = 200 \text{ V} \sin(\omega t + \frac{\pi}{6})$.
  Peak Voltage ($V_m$) = $200 \text{ V}$. Phase $\phi_v = \frac{\pi}{6} = 30^\circ$.
• Time Period: $T = 20 \text{ ms} = 20 \times 10^{-3} \text{ s}$.
• Angular Frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{20 \times 10^{-3}} = 100\pi \text{ rad/s}$.
• Phase Difference ($\phi$): Voltage leads current by $\phi = 30^\circ$.

2. Evaluate the Options

(a) Current lags the voltage

Voltage phase is $+30^\circ$ relative to current. Therefore, voltage leads current, which means current lags voltage.

Statement (a) is CORRECT.

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(b) Resistance in the circuit is $250\sqrt{3} \, \Omega$

First, calculate the Impedance ($Z$) using the peak values:

$$Z = \frac{V_m}{I_m} = \frac{200 \text{ V}}{0.4 \text{ A}} = 500 \, \Omega$$

The resistance ($R$) is the component of impedance in phase with the current:

$$R = Z \cos \phi = 500 \cos(30^\circ)$$ $$R = 500 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \, \Omega$$

This matches the value in the statement.

Statement (b) is CORRECT.

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(c) Inductance Calculation

The circuit is inductive (Voltage leads). Using the phase tangent formula:

$$\tan \phi = \frac{X_L – X_C}{R}$$ $$\tan(30^\circ) = \frac{X_L – 74}{250\sqrt{3}}$$ $$\frac{1}{\sqrt{3}} = \frac{X_L – 74}{250\sqrt{3}}$$

Multiply both sides by $250\sqrt{3}$:

$$250 = X_L – 74$$ $$X_L = 324 \, \Omega$$

Now, calculate Inductance ($L$):

$$L = \frac{X_L}{\omega} = \frac{324}{100\pi} \approx 1.03 \text{ H}$$

The statement claims $560 \text{ mH}$, which is incorrect.

Statement (c) is INCORRECT.

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(d) Average power dissipation in the circuit is $20\sqrt{3} \text{ W}$

Average power ($P$) is given by:

$$P = V_{rms} I_{rms} \cos \phi = \frac{V_m I_m}{2} \cos \phi$$ $$P = \frac{200 \times 0.4}{2} \cos(30^\circ)$$ $$P = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ W}$$

This matches the value in the statement.

Statement (d) is CORRECT.

Final Answer

The correct statements are (a), (b), and (d).