In the above circuit, $$C = \frac{\sqrt{3}}{2}\mu F$$, $$R_2 = 20 \Omega$$, $$L = \frac{\sqrt{3}}{10} H$$ and $$R_1 = 10 \Omega$$. Current in $$L$$-$$R_1$$ path is $$I_1$$ and in C-$$R_2$$ path it is $$I_2$$. The voltage of AC source is given by, $$V = 200\sqrt{2} \sin(100t)$$ volts. The phase difference between $$I_1$$ and $$I_2$$ is:

We first note that the source voltage is written as $$V = 200\sqrt{2}\,\sin(100t)$$. Comparing this with the standard form $$V = V_{0}\sin(\omega t)$$ we can at once read the angular frequency as $$\omega = 100\;{\rm rad\,s^{-1}}.$$

There are two independent series branches connected across this same source.

Branch 1 (R-L branch)

The impedance of a resistor $$R_1$$ in series with an inductor $$L$$ is, by definition,

$$Z_1 = R_1 + j\omega L,$$

where the symbol $$j$$ stands for $$\sqrt{-1}$$ (conventional electrical engineering notation).

Substituting the numerical data

$$R_1 = 10\;\Omega,\qquad L = \frac{\sqrt{3}}{10}\;{\rm H},\qquad \omega = 100\;{\rm rad\,s^{-1}},$$

we obtain

$$\omega L \;=\; 100 \times \frac{\sqrt{3}}{10}\;=\;10\sqrt{3}\;\Omega.$$

Hence

$$Z_1 = 10 + j\,10\sqrt{3}\;\Omega.$$

The phase angle by which the branch current lags its voltage for an R-L circuit is given by the well-known relation

$$\phi_1 = \tan^{-1}\!\Bigl(\frac{\omega L}{R_1}\Bigr).$$

Substituting,

$$\phi_1 \;=\; \tan^{-1}\!\Bigl(\frac{10\sqrt{3}}{10}\Bigr) \;=\; \tan^{-1}(\sqrt{3}) \;=\; 60^{\circ}.$$

Because the term $$+j\omega L$$ is inductive, this angle represents lag. Therefore

$$I_1 \;\text{lags}\; V \;\text{by}\; 60^{\circ},$$

or equivalently, the phase of $$I_1$$ is

$$\angle I_1 = -60^{\circ}.$$

Branch 2 (R-C branch)

The impedance of a resistor $$R_2$$ in series with a capacitor $$C$$ is

$$Z_2 = R_2 - \dfrac{j}{\omega C}.$$

Using the given values

$$R_2 = 20\;\Omega,\qquad C = \frac{\sqrt{3}}{2}\,\mu{\rm F} = \frac{\sqrt{3}}{2}\times 10^{-6}\;{\rm F},$$

we first compute the capacitive reactance $$X_C$$:

$$\omega C = 100 \times \frac{\sqrt{3}}{2}\times 10^{-6} \;=\; 50\sqrt{3}\times 10^{-6}.$$

Taking the reciprocal,

$$X_C \;=\; \frac{1}{\omega C} \;=\; \frac{1}{50\sqrt{3}\times 10^{-6}} \;=\; \frac{10^{6}}{50\sqrt{3}} \;=\; \frac{20\,000}{\sqrt{3}}\;\Omega \;=\; \frac{20\,000\sqrt{3}}{3}\;\Omega.$$

For later numerical clarity we also write

$$X_C \;\approx\; 11\,546.7\;\Omega.$$

Thus

$$Z_2 = 20 \;-\; j\,X_C = 20 - j\,11\,546.7\;\Omega.$$

The phase angle of the current with respect to the voltage for an R-C series circuit is given by

$$\phi_2 = \tan^{-1}\!\Bigl(\frac{X_C}{R_2}\Bigr),$$

and because the reactance term carries a minus sign, this angle represents a lead (capacitive currents lead the voltage).

Substituting,

$$\phi_2 \;=\; \tan^{-1}\!\Bigl(\frac{11\,546.7}{20}\Bigr) \;=\; \tan^{-1}(577.335) \;\approx\; 89.9^{\circ}.$$

To a very good approximation we may treat this as exactly

$$\angle I_2 = +90^{\circ}\;,$$

i.e. the current $$I_2$$ leads the source voltage by $$90^{\circ}.$$

Phase difference between the two branch currents

The current $$I_1$$ has phase $$-60^{\circ}$$, while $$I_2$$ has phase $$+90^{\circ}$$. The absolute phase difference $$\Delta\phi$$ is therefore

$$\Delta\phi = (+90^{\circ}) - (-60^{\circ}) = 90^{\circ} + 60^{\circ} = 150^{\circ}.$$

This is the required result.

Hence, the correct answer is Option D.