Two resistor are connected in series across a 5 V rms source of alternating potential. The potential difference across `6 Omega` resistor is 3V. If R is replaced by a pure inductor L of such magnitude that current reamins same. Then the pontential difference across L is

Options

A. (A) 1 V
B. (B) 2 V
C. (C) 3 V
D. (D) 4 V

Answer

D

Explanation

`V_(6 Omega)=3 =6(I_V) :. (I_V) = 0.5A`
`(I_V)=1/2 = (5)(sqrt(6^(2)+X_(L)^(2))), (X_L)=8 Omega` ltbRgt Now, `(V_L) = (I_V)*(X_L) =1/2 xx 8 = 4 V`.