An alternating emf of 110 V is applied to a circuit containing a resistance R of 80 Ω and an inductor L in series. The current is found to lag behind the supply voltage by an angle 8 = tan^{-1} (3/4). Find the :

(i) Inductive reactance

(ii) Impedance of the circuit

(iii) Current flowing in the circuit

(iv) If the inductor has a coefficient of self-inductance of 0.1 H, what is the frequency of the applied emf?

#### Solution

(i) Here, E_{v} = 110 V, R = 80 Ω, L = 0.1 H

θ = tan^{-1}`(3/4)`

We know that in an L - R circuit

tan θ = `"Lω"/"R" = "X"_"L"/"R"`

`3/4 = "X"_"L"/80`

`"X"_"L" = 80 x 3/4` = 60 Ω

(ii) impedance Z = `sqrt( "X"_L^2 + "R"^2 )`

Z = `sqrt( (60)^2 + (80)^2 )`

Z = 100 Ω

(ii) Now,

`"I"_"v" = "E"_"v"/"Z" = 110/100 = 1.1 "A"`

(iv) X_{L} = 2πnL

∴ n = `"X"_"L"/(2π"L") = 60/( 2 xx 3.14 xx 0.1 ) = 95.54 "Hz"`.