The power factor of a series RC circuit is 1/√2. If the frequency of the applied AC is halved, the power factor will be 1/√5.
Given that the power factor of the series RC circuit is 1/√2.
For series RC circuit power factor
cos ∅ =R/Z=R/√(R2+XC2)=1/√2
When the frequency is halved
XC is doubled
Now the power factor
cos ∅ =R/Z=R/√(R2+XC'2)=R/√(R2+(2R)2)=1/√5
Now power factor of the RC circuit is 1/√5.
Hence option C is the correct answer.