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

Two resistor are connected in series across a 5 V rms source of altern

1.	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 A. (A) 1 VB. (B) 2 VC. (C) 3 VD. (D) 4 V
Answer» Correct Answer - D `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`.

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