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AC Reactive Circuit Calculations
26 questions
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Question 1 of 26
Determine the input frequency necessary to give the output voltage a phase shift of 75$^{o}$:
Also, write an equation that solves for frequency ($f$), given all the other variables ($R$, $L$, and phase angle $\theta$).
Reveal answerKey Application for this Solution: RL Filter Phase Angle Equation
The fundamental phase angle equation relates resistance to induction in the following form:
$$ \tan \theta = {X_L \over R}$$
Since inductive reactance $X_L$ is equal to $2 \pi fL$ then the substituted phase angle equation is then:
$$ \tan \theta = {2 \pi fL \over R}$$
Solving this equation for frequency yields:
$$f = {R \tan \theta \over {2 \pi L}}$$
Finally, using the values provided, the actual frequency yields:
$$f = 157.970 kHz$$
Notes:Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.
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Question 2 of 26
An AC electric motor under load can be considered as a parallel combination of resistance and inductance:
Calculate the equivalent inductance ($L_{eq}$) if the measured source current is 27.5 amps and the motor’s equivalent resistance ($R_{eq}$) is 11.2 $\Omega$.
Reveal answerKey Application for this Solution: Inductive Load Effects
In AC parallel circuits, branch currents add in phasor fashion, which means the inductive current and resistive current form two sides of a right triangle. The resistive current may be calculated using Ohm’s Law ($I = {V \over R} = {277 \hbox{ V} \over 11.2 \> \Omega} = 24.73$ A):
The inductive current, therefore, may be calculated using the Pythagorean Theorem:
$${I_{total}}^2 = {I_R}^2 + {I_L}^2$$
$${I_L}^2 = {I_{total}}^2 - {I_R}^2$$
$$I_L = \sqrt{{I_{total}}^2 - {I_R}^2}$$
$$I_L = \sqrt{(27.5 \hbox{ A})^2 - (24.73 \hbox{ A})^2} = 12.02 \hbox{ A}$$
Knowing that the inductive current is 12.02 amps allows us to use Ohm’s Law to calculate the inductive reactance:
$$X = {V \over I}$$
$$X = {277 \hbox{ V} \over 12.02 \hbox{ A}} = 23.04 \> \Omega$$
Now we may calculate inductance from inductive reactance:
$$X_L = 2 \pi f L$$
$$L = {X_L \over 2 \pi f}$$
$$L = {23.04 \> \Omega \over (2 \pi) (60 \hbox{ Hz})} = 61.11 \hbox{ mH}$$
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Question 3 of 26
An AC electric motor under load can be considered as a parallel combination of resistance and inductance:
Calculate the current necessary to power this motor if the equivalent resistance and inductance is 20 $\Omega$ and 238 $mH$, respectively.
Reveal answerKey Application for this Solution: Inductive Load Effects
In a parallel R-L circuit, the currents are added as the legs of a right triangle.
The resistance yields a current following Ohm’s Law:
$$I_R = {V \over R}$$
$$I_R = {240V \over 20 \Omega}$$
$$I_R = 12.0 A$$
Reactive current involves first determining the inductive reactance, then using Ohm’s Law:
$$X_L = 2 \pi f L$$
$$X_L = (2 \pi)(60Hz)(238mH)$$
$$X_L = 89.7 \Omega$$
Now use Ohm’s Law:
$$I_L = {V \over R}$$
$$I_L = {240V \over 89.7 \Omega}$$
$$I_L = 2.68 A$$
Finally, use the Pythagorean Equation to add the vector sums of the currents:
$$I_{supply} = \sqrt{(I_R)^2 + (I_L)^2}$$
$$I_{supply} = \sqrt{(12.0 A)^2 + (2.68 A)^2}$$
$$I_{supply} = 12.29 A$$
Notes:This is a practical example of a parallel LR circuit, as well as an example of how complex electrical devices may be “modeled” by collections of ideal components. To be honest, a loaded AC motor’s characteristics are quite a bit more complex than what the parallel LR model would suggest, but at least it’s a start!
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:
Step 1: Calculate all reactances ($X$).
Step 2: Draw an impedance triangle ($Z$ ; $R$ ; $X$), solving for $Z$
Step 3: Calculate circuit current using Ohm’s Law: $I = {V \over Z}$
Step 4: Calculate series voltage drops using Ohm’s Law: $V = {I Z}$
Step 5: Check work by drawing a voltage triangle ($V_{total}$ ; $V_1$ ; $V_2$), solving for $V_{total}$
By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.
