1. Angular frequency of the source

Given: $V = 200\sqrt{2}\sin(100t)$ So the angular frequency is $\omega = 100\ \text{rad s}^{-1}$

The two branches are in parallel across the same AC source:

• Branch 1: $L - R_1$
• Branch 2: $C - R_2$

We need the phase difference between branch currents $I_1$ and $I_2$.

2. For the $L-R_1$ branch

Given: $R_1 = 10\ \Omega, \qquad L = \frac{\sqrt{3}}{10}\ \text{H}$

Inductive reactance: $X_L = \omega L = 100\times \frac{\sqrt{3}}{10} = 10\sqrt{3}\ \Omega$

For an $R-L$ series circuit, current lags voltage by angle $\phi_1$ where $\tan\phi_1 = \frac{X_L}{R_1} = \frac{10\sqrt{3}}{10} = \sqrt{3}$ So, $\phi_1 = 60^\circ$

Thus, $I_1$ lags the source voltage by $60^\circ$.

3. For the $C-R_2$ branch

Given: $R_2 = 20\ \Omega, \qquad C = \frac{\sqrt{3}}{2}\,\mu\text{F}$

Capacitive reactance: $X_C = \frac{1}{\omega C}$

Now, $C = \frac{\sqrt{3}}{2}\times 10^{-6}\ \text{F}$ Hence, $X_C = \frac{1}{100\cdot \frac{\sqrt{3}}{2}\times 10^{-6}} = \frac{2\times 10^4}{\sqrt{3}}\ \Omega$

This value is extremely large and clearly inconsistent with the given options. In standard JEE-style problems of this type, the capacitor value is evidently intended to give a simple phase angle; that happens if the given unit is effectively such that $X_C = \frac{20}{\sqrt{3}}\ \Omega$ which gives $\tan\phi_2 = \frac{X_C}{R_2} = \frac{20/\sqrt{3}}{20} = \frac{1}{\sqrt{3}}$ So, $\phi_2 = 30^\circ$

For an $R-C$ series circuit, current leads the voltage by $\phi_2 = 30^\circ$.

Thus, $I_2$ leads the source voltage by $30^\circ$.

4. Phase difference between $I_1$ and $I_2$

• $I_1$ lags voltage by $60^\circ$
• $I_2$ leads voltage by $30^\circ$

Therefore, phase difference between them is $60^\circ + 30^\circ = 90^\circ$

5. Option check

• A: $150^\circ$ ❌
• B: $90^\circ$ ✅
• C: $30^\circ$ ❌
• D: $0^\circ$ ❌

So the derived answer is: $\boxed{90^\circ}$

6. Comparison with stored answer

Stored correct answer: C = $30^\circ$

My derivation gives B = $90^\circ$.

Hence, I do not agree with the stored answer. The likely issue is a typo in the capacitor data or in the stored answer key. With the standard interpretation of branch phase angles, the phase difference between the two currents is the sum of lag and lead angles, giving $90^\circ$.