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Dear student, I am attaching the solution of above question below. Thank youb) Case-2 Solving Case- II 8irst\rightarrow Original p f=\cos \phi_{1}=0.8 \mathrm{lag}\rightarrow final p f=\cos \phi_{2}= Unity\rightarrow Power P=200 \mathrm{~W}\rightarrow \phi_{1}=\cos ^{-1}(0.8)=36.86^{\circ}\rightarrow \phi_{2}=\cos ^{-1}(1)=0^{\circ}\rightarrow leading KVAr taken by capacitor bank,\begin{array}{l}=P\left(\tan \phi_{1}-\tan \phi_{2}\right) \\=200\left[\tan (36.86)-\tan \left(0^{\circ}\right)\right] \\=200 \times[0.749-0] \\=149.8 \text { KVAr }\end{array}\rightarrow leading KVAr taken by each three set,=\frac{149.8}{3}=49.33 \mathrm{KVAr}\rightarrow Now, let C_{\text {ph be the value of }} Capacitance,\rightarrow Phase current of capacitor given by,\begin{aligned}I_{c p} & =\frac{V_{p h}}{x_{c}}=2 \pi f C V_{p h} \\& =2 \pi \times 50 \times C_{p h} \times 400\end{aligned}I_{c p}=40000 \pi C_{p h} AmperesNow,\begin{array}{l}\text { KVAr/phase }=\frac{V_{p h} I_{c p}}{1000}=\frac{400 \times 40000 \pi \mathrm{Cph}}{1000} \\\frac{49.33 \text { VAr }}{1000}=\frac{400 \times 4000 \pi \mathrm{cph}}{1000} \\C_{p h}=9.93 \times 10^{-4} \text { Farad needed to be comect per phase }\end{array}C_{p h}=9.93 \times 10^{-4} Farad \overrightarrow{\text { needed to be consec }} per phasea) Case = 1In Case the Capacitors are connected in star across the 3-\phi line than the value of Capacitor is \frac{1}{3} times of the value of Capacitance when connected in \triangle, So,\begin{array}{l}C_{Y}=\frac{1}{3} C_{\Delta} \quad[\text { for pf correction] } \\C_{Y}=\frac{1}{3} \times 9.93 \times 10^{-4} \text { Farad } \\C_{Y}=3.31 \times 10^{-4}\end{array} ...