A) 31 \[\mu F\]
B) 54 \[\mu F\]
C) 151 \[\mu F\]
D) 201 \[\mu F\]
Correct Answer: B
Solution :
Ist Case: From formula \[R=\frac{{{V}^{2}}}{P}\] \[=\frac{110\times 110}{330}\] \[=\frac{110}{3}\,\Omega \] Since, current lags the voltage thus, the circuit contains resistance and inductance. Power factor \[\cos \phi =0.6\] \[\frac{R}{\sqrt{{{R}^{2}}+X_{L}^{2}}}=0.6\] \[\Rightarrow \] \[{{R}^{2}}+X_{L}^{2}={{\left( \frac{R}{0.6} \right)}^{2}}\] \[\Rightarrow \] \[X_{L}^{2}=\frac{R}{{{(0.6)}^{2}}}-{{R}^{2}}\] \[\Rightarrow \] \[X_{L}^{2}=\frac{{{R}^{2}}\times 0.64}{0.36}\] \[\therefore \] \[{{X}_{L}}=\frac{0.8R}{0.6}=\frac{4R}{3}\] ?. (i) IInd Case : Now \[\cos \phi =1\] (given) therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which \[{{X}_{L}}={{X}_{C}}\] \[\therefore \] \[{{X}_{C}}=\frac{4R}{3}=\frac{4}{3}\times \frac{110}{3}=\frac{440}{9}\Omega \] [ from Eq. (i)] or \[\frac{1}{2\pi fC}=\frac{440}{9}\Omega \] \[\therefore \] \[C=\frac{9}{2\times 3.14\times 60\times 440}\] = 0.000054 F = 54 \[\mu F\]
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