### How Much Lag?

If you know the ratio of the inductive reactance to the resistance (XL/R) in an RL circuit, then you can find the phase angle. Of course, you can also find the phase angle if you know the actual values of XL and R.

#### Pictorial Method

It isn’t necessary to construct an entire RL plane to find phase angles. You can use a ruler that has centimeter (cm) and millimeter (mm) markings, and a protractor. First, draw a line a little more than 10 cm long, going from left to right on a sheet of paper. Use the ruler and a sharp pencil. Then, with the protractor, construct a line off the left end of this first line, going vertically upward. Make this line at least 10 cm long. The horizontal line, or the one going to the right, is the R axis of a coordinate system. The vertical line, or the one going upward, is the XL axis.

If you know the values of XL and R, divide them down or multiply them up so they’re both between 0 and 100. For example, if XL = 680 Ω and R = 840 Ω, you can divide them both by 10 to get XL = 68 and R = 84. Plot these points lightly by making hash marks on the vertical and horizontal lines you’ve drawn. The R mark in this example will be 84 mm to the right of the origin, and the XL mark will be 68 mm up from the origin. Next, draw a line connecting the two hash marks, as shown in above figure. This line will run at a slant, and will form a triangle along with the two axes. Your hash marks, and the origin of the coordinate system, form the three vertices of a right triangle. The triangle is called right because one of its angles is a right angle (90°). Measure the angle between the slanted line and the R axis. Extend one or both of the lines if necessary in order to get a good reading on the protractor. This angle will be between 0 and 90°, and represents the phase angle in the RL circuit.

The complex impedance vector, R + jXL, is found by constructing a rectangle using the origin and your two hash marks as three of the four vertices, and drawing new horizontal and vertical lines to complete the figure. The vector is the diagonal of this rectangle, as shown in following figure. The phase angle is the angle between this vector and the R axis. It will be the same as the angle of the slanted line in above figure. #### Trigonometric Method

If you have a good scientific calculator that can find the arctangent of a number (also called the inverse tangent and symbolized either as arctan or tan−1), you can determine the RL phase angle more precisely than the pictorial method allows. Given the values of XL and R, the RL phase angle is the arctangent of their ratio. Phase angle is symbolized by the lowercase Greek letter phi (pronounced “fie” or “fee” and written φ). Therefore: φ = tan−1 (XL/R) or φ = arctan (XL/R)

#### Problem 1

Suppose the inductive reactance in an RL circuit is 680 Ω and the resistance is 840 Ω. What is the phase angle?
The ratio XL/R is 680/840. A calculator will display this quotient as something like 0.8095 and some more digits. Find the arctangent of this number. You should get 38.99 and some more digits. This can be rounded off to 39.0°.

#### Problem 2

Suppose an RL circuit operates at a frequency of 1.0 MHz with a resistance of 10 Ω and an inductance of 90 μH. What is the phase angle? What does this tell us about the nature of this RL circuit at this frequency?
Find the inductive reactance using the formula XL = 6.28fL = 6.28 × 1.0 × 90 = 565 Ω. Then find the ratio XL/R = 565/10 = 56.5. The phase angle is equal to arctan 56.5, which, rounded to two significant figures, is 89°. The circuit contains an almost pure inductive reactance, because the phase angle is close to 90°. The resistance contributes little to the behavior of this RL circuit at 1.0 MHz.

#### Problem 3

What is the phase angle for the preceding circuit at a frequency of 10 kHz? With that information, what can we say about the behavior of the circuit at 10 kHz?
This requires that XL be calculated again, for the new frequency. Let’s use megahertz, so it goes in the formula with microhenrys. A frequency of 10 kHz is the same as 0.010 MHz. Calculating, we get XL = 6.28fL = 6.28 × 0.010 × 90 = 5.65 Ω. The ratio XL/R is 5.65/10 = 0.565. Therefore, the phase angle is arctan 0.565, which, rounded to two significant figures, is 29°. This is not close to either 0° or 90°. Thus, at 10 kHz, the resistance and the inductive reactance both play significant roles in the behavior of the circuit.