What is LR Circuit: Derivation and Its Diagram

What is the Power Dissipated by the L-R circuit?

Fig. shows the circuit.

The applied alternating voltage,
V = V0sinωt
If the instantaneous change in current in the circuit be dI, then the induced emf in the inductor L = –dIdt;
i.e., the effective voltage in the circuit,
Ve = V0sinωt – LdIdt
According to Ohm’s law, Ve = IR, where the resistance, if any, of the inductor L is included in R.
∴ V0sinωt – LdIdt = IR
or, LdIdt + RI = V0ωt …. (2)
Let I = I0sin(ωt + α)
or, dIdt = ωI0 cos(ωt + α)
So, putting the values of I and dIdt identity (2) we get,
ωLI0cos(ωt + α) + RI0 sin(ωt + α) = V0sinωt
or I0{Rsin(ωt + α) + ωLcos(ωt + α)} = V0sinωt

i) The current lags behind the applied voltage by a phase angle θ given by tan θ = sinθ = ωLZ×ZR = ωLR
i.e., θ = tan-1(ωLR).

This phase relation is shown in Fig. We know that, in a pure resistive circuit V and I are in the same phase i.e., phase difference, θ = 0. On the other hand, in a pure inductive circuit, I
always lags behind V by θ = 90°. So in an LR circuit, I should lag behind V by 0< θ < 90°.

ii) In this circuit Z plays the same role as R in a pure resistive circuit. Z is known as the impedance of an ac circuit.
Impedance, Z = R2+(ωL)2−−−−−−−−−√ = R2+(XL)2−−−−−−−−−√ …… (4)
where, XL = ωL = inductive reactance
Impedance in an LR circuit is the effective resistance of the circuit arising from the combined effects of ohmic resistance and inductive reactance.

We can express the relation among R, XL and Z with a suitable right angled triangle [Fig.]. This triangle is called the impedance triangle.

Note that, R, ωL and Z have the same unit ohm (Ω).

Power in the circuit:

Here, power factor of the circuit, cosθ = RZ[Fig.].
∴ P = 12V0Iocosθ = 12(I0Z)I0⋅RZ = 12I20R = (I02√)2R
i.e., P = I2rmsR
This means that, the power is dissipated only in the resistance R. Current through the inductor is wattless.