Question A 400 V, 50 Hz, three phase line delivers 200 kW at 0.8 lagging power factor. It is desired to raise the line power factor to unity by installing shunt capacitors. Calculate the capacitance of each unit if they are connected in, a) Star (as given in Fig. 4 (a)) b) Delta (as given in Fig. 4 (b)) HH Three phase load Three phase load Fig. 04 (a) Fig. 04(b)

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} ...