# The LR (inductance-resistance) circuit

The circuit in Figure
1 contains both inductance and resistance. As with the capacitance-resistance circuit, the
current through it depends on the value of both the components and the frequency of the supply
voltage.

Let the supply voltage
be v

_{0} and the voltages across the inductor and the resistor be v

_{L0} and v

_{R0} respectively. Now we
know that for a resistor the current and voltage are in phase, while for an inductor the current
leads the voltage by 90

^{o}; V

_{L0} therefore leads v

_{0} by 90

^{o}, as can be seen from Figure 2.

The resultant voltage is
given by

v

_{0}^{2}=
v

_{R}_{0}^{2} + v

_{L}_{0}^{2} =
i

_{0}^{2}R

^{2} + i

_{0}^{2}X

_{L}^{2}The current in
the circuit is therefore: i

_{o} =v

_{o}/[X

_{L}^{2} + R

^{2}]

^{1/2 }
and the impedance (Z) is:

Z = [X_{L}^{2} + R^{2}]^{1/2} = [w^{2}L^{2} + R^{2}]^{1/2}
The phase angle for this circuit (

w) (see Figure 3) is given by

tan f = v_{L0}/v_{R}0 = wL/R