Phase difference between I₂ and V, i.e. C $$-$$ R₂ circuit is given by

$$\tan \phi = {{{X_C}} \over {{R_2}}} \Rightarrow \tan \phi = {1 \over {C\omega {R_2}}}$$

Substituting the given values, we get

$$\tan \phi = {1 \over {{{\sqrt 3 } \over 2} \times {{10}^{ - 6}} \times 100 \times 20}} = {{{{10}^3}} \over {\sqrt 3 }}$$

$$\therefore$$ $$\phi$$₁ is nearly 90$$^\circ$$.

Phase difference between I₁ and V, i.e. in L $$-$$ R₁ circuit is given by

$$\tan {\phi _2} = - {{{X_L}} \over {{R_1}}} = - {{L\omega } \over R}$$

Substituting the given values, we get

$$\tan {\phi _2} = - {{{{\sqrt 3 } \over {10}} \times 100} \over {10}} = - \sqrt 3 $$

As, $$\tan {\phi _2} = - \sqrt 3 $$

$$\therefore$$ $${\phi _2} = 120^\circ $$

Now, phase difference between I₁ and I₂ is

$$\Delta \phi = {\phi _2} - {\phi _1} = 120^\circ - 90^\circ = 30^\circ $$