A) \[31\text{ }\mu F\]
B) \[\text{54 }\mu F\]
C) \[\text{151 }\mu F\]
D) \[\text{201 }\mu F\]
Correct Answer: B
Solution :
Resistance of circuit, \[R=\frac{{{V}^{2}}}{P}=\frac{110\times 110}{330}=\frac{110}{3}\Omega \] 1st case: Power factor\[\cos \phi =0.6\] Since, current lags the voltage thus, the circuit contains resistance and inductance. \[\therefore \] \[\cos \phi =\frac{R}{\sqrt{{{R}^{2}}+X_{L}^{2}}}=0.6\] Or \[{{R}^{2}}+X_{L}^{2}={{\left( \frac{R}{0.6} \right)}^{2}}\] Or \[X_{L}^{2}=\frac{{{R}^{2}}}{{{(0.6)}^{2}}}-{{R}^{2}}\] Or \[X_{L}^{2}=\frac{{{R}^{2}}\times 0.64}{0.36}\] \[\therefore \] \[{{X}_{L}}=\frac{0.8R}{0.6}=\frac{4R}{3}\] ?(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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