In a series LR circuit, power of 400W is dissipated from a source...

$P = {V}_{m}$. in $cosϕ$

$400 = 250 \times 1m \times 0 .8$

${i}_{rms} = 2A$

${\left(1m\right)}^{2} . R = P$

$4 \times R = 400$

$⇒$ $R = 100Ω$.

$cosϕ = \dfrac{R}{\sqrt{{R}^{2} + {X}_{L}^{2}}}$

${100}^{2} + {X}_{L}^{2} = {\left(\dfrac{100}{0 .8}\right)}^{2}$

${100}^{2} + {X}_{L}^{2} = {\left(\dfrac{100}{0 .8}\right)}^{2}$

${X}_{L} = 75Ω$

Power factor is unity

${X}_{C} = {X}_{L} = 75$

$\dfrac{1}{\omega} = 75$

$⇒$ $C = \dfrac{1}{75 \times 2H \times 50} = \dfrac{1}{7500\pi}F$

$3\pi \times 2500$

$= \dfrac{1}{3\pi} \times 4 \times {10}^{2}mF$

$= \dfrac{400}{3\pi}\mu F$

$N = 400$