Wires A and B have resistivities `p_A` and `p_B`. If the diameter of the wire B is twice that of A and the Two wires have same resistance, then `(L_b)/(L_A)` is

Options

A. 2
B. 1
C. `(1)/(2)`
D. `(1)/(4)`

Answer

B

Explanation

V= 50 sin (50t), `I=50 sin ((50t+(pi)/(3))`
On comparing with the standard equation , we get ,
`E_circ = 50, i_circ = 50,phi= (pi)/(3)`
Power dissipated , `P = (epsilon_circi_circ )/(2)cos phi`
`Rightarrow P= (50xx50)/(2)cos ((pi)/(3))`
`therefore P= 1.25W `