How to solve rl circuits

How To Calculate Rl Series Circuits

The presence of an inductor in a series circuit causes the voltage and current to be out of phase. Maximum voltage leads maximum current by 90 in inductors. Maximum voltage and current are in phase, 0, in resistors. An inductor and resistor in series will have voltage and current out of phase by an angle between 0 and 90. The angle will depend on the inductive reactance of the inductor and the value of the resistor. The vector diagram in figure one illustrates how the phase angle,  is related to inductive reactance and resistance.

The total opposition to current flow in AC circuits is called impedance. Impedance is identified using the letter “Z”. Impedance, Z is calculated by adding inductive reactance, XL, the value of resistance, R, at 90. Trigonometry or the Pythagorean Theorem can be used to make this calculation. A fully worked out series RL circuit analysis can be seen by clicking on the example tab at the upper left of this page.

Click on the “Begin” button when you are ready to start the exercise. Each use of the activity will have new component values and voltages assigned. The exercise has a help option that will give formulas and procedures for analyzing the circuit. A small score penalty is applied for each use of the help option. One should practice without concern for scores until the solution procedure has been mastered. You should then attempt to get high scores in minimum time without the using the help option. Keep in mind that there may be more than one approach to analyzing the circuit.

Find XL, Z, IR, IL, IT, VR, VL, PT and PA for the circuit in figure one.

XL, inductive reactance, is the opposition to current flow produced by inductors in AC circuits.

XL = 2f L = (2)(3.14159)(150)(1.06) = 999

Z, impedance, is the total opposition to current flow in an AC circuit. The Z vector is the hypotenuse of a right triangle where the other two sides are the XL vector and the R vector. The Pythagorean Theorem is one way to solve for Z.

__________ _____________

Z =XL2 + R2 = 9992 + 6802 =1208.4

IR, resistor current, is the same as total current because the circuit is series.

IL, inductor current, is the same as total current because the circuit is series.

IT, total current, is the total current delivered to the circuit by the AC source.

V 2.9

IR = IL = IT = ---- = --------- = 2.4mA

Z 1208.4

phase angle, is the angle between maximum voltage and maximum current in an AC circuit. The phase angle is the angle between the Z vector and the R vector on the vector diagram shown in figure two.

XL 999

tan = ------ = ------- = 1.4691

R 680

= 55.757

VR, voltage drop across the resistor, is found by use of Ohm's law.

VR = ( I )( R ) = (2.4mA )( 680 ) = 1.632V

VL, voltage drop across the inductor, is found by use of Ohm's law.

VL = ( I )( XL ) = (2.4mA )( 999 ) = 2.3976V

PT, true power, is found by calculating the resistor power consumption. Resistors are the only circuit property that actually converts electrical energy to heat. Inductors merely store and give energy back to the circuit.

PT = ( IR )( VR ) = (2.4mA )( 1.632V ) = 3.9168mW

PA, apparent power, is the power seemingly being delivered to the circuit by the AC source. The calculation does not result in “true” power because the current and voltage are out of phase. Apparent power is measured in Volt-Amperes, VA, units.

PA = ( IS )( VS ) = (2.4mA )( 2.9003V ) = 6.9607mVA

Several quantities could have been calculated by other methods. Trigonometry could have been used to calculate Z instead of using the Pythagorean Theorem.