The R-L circuit contains resistors and inductors, as shown in Figure 1. If the resistance is very small or equal to zero, the voltage and current waveforms are shown in Figure 2. The power supply voltage has reached the maximum value at the moment when the current has just entered the positive half cycle. The current takes time to build up the magnetic field, and the power supply voltage on the coil has already begun to drop. In a purely inductive circuit, the power supply voltage leads the current by 90°. Note that when the current reaches its maximum value and the magnetic field is built up, the voltage is zero. When the voltage turns negative, the current decreases and the magnetic field gradually collapses. When the resistance in the R-L circuit increases, the phase difference between voltage and current is less than 90°.

The resistance of an inductor to current is called inductive reactance X_L, and the unit of measurement is ohms. The calculation formula of inductance is X_L=2πfL, where f is frequency; L is inductance; it is equal to 3.14. When calculating impedance, resistance and inductive reactance should be combined, but like capacitive reactance, these two types of values ​​cannot be added directly. The impedance of the R-L series circuit is Z=√(R²+X²_L). If R=20Ω and X_L=40Ω, then Z=√(20²+40²)Ω. Please note that in this formula, X_L will increase with increasing frequency and decrease with decreasing frequency.

①Calculation of R-L series circuit
When applying Ohm’s law, Z can be substituted for R. So I=E/Z, E=I×Z, Z=E/I.

example 1
Using the values ​​in Figure 3, calculate Z, I, V_R, and VX_L, calculate the sum of the voltage drops of each component, and confirm that it is equal to the power supply voltage.
Impedance is
Z=√(R²+X²L)=√(50²+30²)Ω=58.31Ω
Apply Ohm’s Law
I_T=E_T/Z=120/58.31A=2.05A
V_R=I×R=2.05×50V=103V
Vx_L=I×X_L=2.05×30V=61.5V
The sum of the pressure drops is as follows
E=√(V²_R+V²x_L)=√(103²+31.5²)V=120V

Example 2
For the circuit shown in Figure 4, when L=79.6mH, calculate the frequency of the voltage source.
The finishing formula X_L=2πfL, we get
F=X_L/2πL=30/(2×3.14×0.0796)Hz=60)Hz
If the frequency is increased to 120 Hz, calculate X_L.
X_L=2πfL=2×3.14×120×0.0796Ω=60Ω

As the frequency increases, the inductive reactance increases, so the current in the circuit will decrease; if the frequency decreases, the current will increase. In some applications, such as electric motors on airplanes, the current will decrease when the frequency of the power supply is increased, and the amount of heat generated will also decrease. These will help reduce the weight of the motor.

②Calculation of R-L parallel circuit
The formula for calculating the total impedance in the R-L parallel circuit is
Z=R×X_L/√(R²+X²L)

Example 3
For the parallel circuit shown in Figure 4, calculate X_L, Z, V_R, V_C, I_R, Ix_L, and I_T.
X_L=2πfL=2×3.14×2000×0.0048Ω=60Ω
Impedance is
Z=R×X_L/√(R²+X²_L)=30×60/√(30²+60²)Ω=1800/67.1Ω=26.8Ω
The voltage on each branch is equal:
V_R=V_X=E=240V
Use Ohm’s law to calculate I_R and Ix_L:
I_R=V_R/R=240/30A=8A
Ix_L=Vx_L/X_L=240/60A=4A
I_T is
I_T=E_T/Z=240/26.8A=8.95A
or
I_T=√(I²_R+I²x_L)=√(8²+4²)A=8.9A
(Either way, the slight difference in the answer comes from the rounding error)

Example 4
In Figure 4, if the frequency becomes 1kHz, calculate X_L, Z, V_R, V_C, I_R, Ix_L, and I_T.
X_L=2πfL=2×3.14×1000×0.0048Ω=30Ω
Impedance is
Z=R×X_L/√(R²+X²_L)=30×30/√(30²+30²)Ω=21.2Ω
The voltage on each branch is equal:
V_R=V_C=E=240V
Calculate the current using Ohm’s law:
I_R=V_R/R=240/30A=8A
I_L=V_L/X_L=240/30A=8A
I_T can be calculated in two ways:
I_T=√(I²_R+I²_L)=√(8²+8²)A=11.3A
or
I_T=E_T/Z=240/21.2A=11.3A