Basic Electricity
Series DC Circuits Practice Worksheet with Answers
27 questions By Tony R. Kuphaldt
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Question 10 of 27
How much voltage does the light bulb receive in this circuit? Explain your answer.

Also, identify the polarity of the voltage across the light bulb (mark with “ ” and “-” signs).
Reveal answer
Follow-up question: being that 30 volts is the commonly accepted “danger” threshold voltage for electric shock, determine whether or not this particular circuit poses a shock hazard.
Notes:This is a very fundamental concept that students must learn: how to determine the total voltage in a series circuit where opposing voltage sources exist. One thing mistake students sometimes make is to try to discern polarity by looking at the polarity signs at the end terminals of the end battery; i.e. at the 3-volt battery’s left-hand terminal, and the 4.5-volt battery’s right-hand terminal, then try to transfer those signs down to the load terminals. This is not an accurate way to tell polarity, but it seems to “work” for them in some situations. This problem is one example of a situation where this faulty technique most definitely does not work!
Have your students collectively agree on a procedure they may use to accurate discern series voltage sums and polarities. Guide their discussion, helping them identify principles that are true and valid for all series circuits.
With regard to the safety question, there is more to determining risk of shock than a simple voltage check. It is important for your students to realize this, despite “accepted” thresholds for hazardous voltage and such.
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Question 11 of 27
Re-draw this circuit in the form of a schematic diagram:

Reveal answer
Notes:One of the more difficult skills for students to develop is the ability to translate the layout of a real-world circuit into a neat schematic diagram. Developing this skill requires lots of practice.
It is very worthwhile for students to discuss how they solve problems such as these with each other. For those students who have trouble visualizing shapes, a simple hint or “trick” to use when translating schematics to illustrations or visa-versa may be invaluable.
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Question 12 of 27
Suppose I connect two resistors in series with one another, like this:

How much electrical resistance would you expect an ohmmeter to indicate if it were connected across the combination of these two series-connected resistors?

Explain the reasoning behind your answer, and try to formulate a generalization for all combinations of series resistances.
Reveal answer
Follow-up question: how much resistance would you expect the ohmmeter to register if there were three similarly-sized resistors connected in series instead of two? What if there were four resistors?
Notes:The concept of series (total) resistance, in relation to individual resistances, usually does not present any difficulties to new students. Parallel resistances are a bit trickier, though . . .







llaboutcircuits.com/worksheets/series-dc-circuits/. In question 5, total series resistance is sum of each resistor, not product. Correct both question and answer
Hello Everybody and thanks for your fantastic articles!
I’m having troubles solving Excercise 19 of your Series circuits problems (https://www.allaboutcircuits.com/worksheets/series-dc-circuits/): For this excercise I’m getting a power dissipation for the heater of 403 W instead of 321.1 W.
Since I’m learning electronics I’m surely doing something wrong… Here how I’m reasoning:
1. With 110 V and a power dissipated of 500 W, I can calculate the current in the circuit, that should be I=P/V 500/110 = 4.545 A
2. Knowing that the current in the circuit is 4.545 A I can use Ohm’s law to calculate the resistance of the heater: R=V/I 110/4.545 = 24 Ohm
3. Now, longing the wires it’s the same as adding 6 Ohm of resistance, for a total resistance of 30 Ohm (24 + 6) so the current in the circuit will change accordingly. To find the new amount of current I use Ohm’s law once again: I=V/R 110/30 = 3.666 A
4. Having a new current value of 3.666 A, I can now calculate power dissipation P=V*I 110*3.666 = 403 W
Could you please help me understand where I’m wrong?
Thank you so much!
Stefano.