A) \[31\,\mu F\]
B) 54\[54\,\mu F\]
C) \[151\,\mu F\]
D) \[201\,\mu F\]
Correct Answer: B
Solution :
Resistance of circuit \[\operatorname{R}=\frac{{{V}^{2}}}{P}=\frac{110\times 110}{330}=\frac{110}{3}\Omega \] Current lags the voltage, it is R - L circuit \[\operatorname{Cos}\phi =0.6=\frac{R}{{{R}^{2}}+{{X}^{2}}_{L}}\] Squaring both side \[0.36=\frac{R}{{{R}^{2}}+{{X}^{2}}_{L}}\] \[0.36={{R}^{2}}+0.36{{X}^{2}}_{\operatorname{L}}={{R}^{2}}\] \[0.36{{X}^{2}}_{\operatorname{L}}={{R}^{2}}-0.36{{R}^{2}}\] \[{{\operatorname{X}}^{2}}_{L}=\frac{4R}{3}\] ?.(1) Now addition of capacitor will result in power factor to unity \[\operatorname{Cos}\phi =\frac{R}{\operatorname{Z}}\] \[\operatorname{Cos}\phi =1\](given) \[\operatorname{R}=Z\] It is condition of resonance \[{{\operatorname{X}}_{L}}={{X}_{C}}\] \[\frac{4R}{3}=\frac{1}{2\pi \operatorname{fc}}\therefore {{\operatorname{X}}_{L}}=\frac{4R}{3}\] \[\frac{4}{3}\times \frac{110}{3}=\frac{1}{2\times 3.14\times 60C}\] \[\operatorname{C}=54\mu F\]
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