Correct Answer - C
For circuit `A`,
Impedance, `Z_(A) = sqrt(R^(2) + (1)/(omega^(2) C^(2)))`
Current in circuit, `I_(A)^(2) = (V)/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))`
Pot. Diff. across `C`,
`V_(C )^(A) = I_(R)^(A) xx (1)/(omega C) = (V)/(sqrt((R omega C)^(2) + 1)`
For circuit `B`, impedance
` Z_(B) = sqrt(R^(2) + (1)/((4 omega)^(2) C^(2)))`
Current in circuit, `I_(B)^(B) = (V)/(sqrt(R^(2) + (1)/((4 omega)^(2) C^(2)))`
Pot. diff. across `C`,
`V_(C )^(B) = I_(R )^(B) xx (1)/( 4 omega C) = (V)/(sqrt((4 R omega C)^(2) + 1))`
We conclude, form (i) and (iii) , `I_(R )^(B) gt I_(R )^(A)`
From (ii) and (iv), `V_(C )^(A) gt V_(C )^(B)` choice (c ) is correct