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Voltage Divider Circuits
37 questions By Tony R. Kuphaldt
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Question 1 of 37
Don’t just sit there! Build something!! Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuit-building exercises, follow these steps:
- Carefully measure and record all component values prior to circuit construction.
- Draw the schematic diagram for the circuit to be analyzed.
- Carefully build this circuit on a breadboard or other convenient medium.
- Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements one-by-one on the diagram.
- Mathematically analyze the circuit, solving for all values of voltage, current, etc.
- Carefully measure those quantities, to verify the accuracy of your analysis.
- If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully re-calculate the values and re-measure.
Avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistors between 1 kΩ and 100 kΩ, unless, of course, the purpose of the circuit is to illustrate the effects of meter loading!
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another time-saving technique is to re-use the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
Reveal answerLet the electrons themselves give you the answers to your own “practice problems”!
Notes:It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.
Students don’t just need mathematical practice. They also need real, hands-on practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to mathematically predict the various voltage and current values. This way, the mathematical theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving equations.
Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, mathematical predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.
Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monologue) format!
A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:
What is the purpose of students taking your course?
If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.
Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.
In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!
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Question 2 of 37
Calculate the output voltages of these two voltage divider circuits (VA and VB):
Now, calculate the voltage between points A (red lead) and B (black lead) (VAB).
Reveal answerVA = + 65.28 V
VB = + 23.26 V
VAB = + 42.02 V (point A being positive relative to point B)
Challenge question: what would change if the wire connecting the two voltage divider circuits together were removed?
Notes:In this question, I want students to see how the voltage between the two dividers’ output terminals is the difference between their individual output voltages. I also want students to see the notation used to denote the voltages (use of subscripts, with an applied reference point of ground). Although voltage is always and forever a quantity between two points, it is appropriate to speak of voltage being “at” a single point in a circuit if there is an implied point of reference (ground).
It is possible to solve for VAB without formally appealing to Kirchhoff’s Voltage Law. One way I’ve found helpful to students is to envision the two voltages (VA and VB) as heights of objects, asking the question of “How much height difference is there between the two objects?”
The height of each object is analogous to the voltage dropped across each of the lower resistors in the voltage divider circuits. Like voltage, height is a quantity measured between two points (the top of the object and ground level). Also like the voltage VAB, the difference in height between the two objects is a measurement taken between two points, and it is also found by subtraction.
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Question 3 of 37
We know that the current in a series circuit may be calculated with this formula:
I = Etotal RtotalWe also know that the voltage dropped across any single resistor in a series circuit may be calculated with this formula:
ER = I R Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only ER expressed in terms of Etotal, Rtotal, and R.
Reveal answerER = Etotal ( R Rtotal) Follow-up question: algebraically manipulate this equation to solve for Etotal in terms of all the other variables. In other words, show how you could calculate for the amount of total voltage necessary to produce a specified voltage drop (ER) across a specified resistor (R), given the total circuit resistance (Rtotal).
Notes:Though this “voltage divider formula” may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one.
Plus signs are missing from these questions and answers, such as in question 11, where the denominator of the voltage divider equation appears as “R1 R2 R3” instead of “R1+R2+R3”. Same in the answer to question 15. Likely elsewhere as well, as Find shows no results for “+” anywhere on the page. I’m using Chrome 88.0.4324.182 on Windows 10, but same on Firefox & Edge. HTML source for the page shows it as “R1 R2 R3”, so it isn’t just a display problem.
Iam a student and i need an explain for this answers how can I get an explanation ?
Ca any one tell in detail how
The formula got formed like this !! ( ER= Et(R/Rt)) thank you