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). - Physics (Theory)

Sum

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, Ev = 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) XL = 2πnL

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

Concept: Induced e.m.f. and Induced Current
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