An AC source is connected to an inductance of 100mH , a capacitance...

From the diagram given,

Angular frequency $\omega = 100rad{s}^{- 1}$ and ${V}_{rms} = \dfrac{{V}_{0}}{\sqrt{2}} = \dfrac{20\sqrt{2}}{\sqrt{2}} = 20V$.

Now, ${X}_{L} = \omegaL = 100 \times 100 \times {10}^{- 3} = 10Ω$ and ${X}_{C} = \dfrac{1}{\omegaC} = \dfrac{1}{100 \times 100 \times {10}^{- 6}} = 100Ω$

\left|\left({X}_{L} - {X}_{C}\right)\right| = \left|10 - 100\right| = 90Ω$</p><p>The impedance of the circuit,</p><p>$Z = \sqrt{{\left({X}_{L} - {X}_{C}\right)}^{2} + {R}^{2}} = \sqrt{{\left(90\right)}^{2} + {\left(120\right)}^{2}} = 150Ω$</p><p>The current in the circuit,</p><p>${i}_{mrs} = \dfrac{{V}_{rms}}{Z} = \left(\dfrac{20}{150}\right)A$</p><p>Power factor$cosϕ = \dfrac{R}{Z}$</p><p>Now</p><p>\left({V}_{rms}{i}_{rms}cosϕ\right)\Deltat = ms\left(\DeltaT\right) ⇒ \Deltat = \dfrac{ms\left(\DeltaT\right)}{{V}_{rms} \times \dfrac{{V}_{rms}}{Z} \times \dfrac{R}{Z}} = \dfrac{ms\left(\DeltaT\right){Z}^{2}}{{{V}_{rms}}^{2} \times R} = \dfrac{2 \times 16 \times 22500}{400 \times 120} = 15s$

$⇒ \Deltat = 15s$