For (p) Insteady state when I = constant

                         V_L = 0 = V₁

So                    V₂ = V

                        Option (C)

For (q) V₁ = 0 again as I = constant

                         V₂ = V

Also V₂ = IR ⇒ Propotional to I.

   Option (B), (C), (D)

For (r) X_L = $\mathrm{\omega }$L = (100 $\mathrm{\pi }$) 6 × 10^–3 ;1.88 $\mathrm{\Omega }$

R = 2$\mathrm{\Omega }$

V₁ = I X_L; V₂ = IR

So V₂ > V₁

V₂ $\propto$ I

also V₁ ∝ I        Option (A), (B), (D)

For (s) V₁ = I X_L

V₂ = I X_C where X_C = $\frac{1}{\mathrm{\omega C}}\simeq 1061\mathrm{\Omega }$

again V₁ $\propto$ I; V₂ $\propto$ I, I ≠ 0

Option (A), (B) (D)

For (t) V₁ = IR when      R = 1000 $\Omega$

V₂ = I X_C when X_C $\simeq$1061 $\mathrm{\Omega }$

V₂ > V₁

V₁, V₂ ∝ I  and I ≠ 0

Option (A), (B), (D)