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Basic Electricity

Series DC Circuits Practice Worksheet with Answers


27 questions By Tony R. Kuphaldt

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  • Question 4 of 27

    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:

    1. Carefully measure and record all component values prior to circuit construction.
    2. Draw the schematic diagram for the circuit to be analyzed.
    3. Carefully build this circuit on a breadboard or other convenient medium.
    4. 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.
    5. Mathematically analyze the circuit, solving for all values of voltage, current, etc.
    6. Carefully measure those quantities, to verify the accuracy of your analysis.
    7. 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.

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  • Question 5 of 27

    The formula for calculating total resistance of three series-connected resistors is as follows:


    R = R1 + R2 + R3



    Algebraically manipulate this equation to solve for one of the series resistances (R1) in terms of the other two series resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.

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  • Question 6 of 27

    Identify which of these circuits is a series circuit (there may be more than one shown!):




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  • B
    beelzebob May 17, 2020

    llaboutcircuits.com/worksheets/series-dc-circuits/. In question 5, total series resistance is sum of each resistor, not product. Correct both question and answer

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  • stefano.salari June 08, 2020

    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.

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    • RK37 June 09, 2020
      The mistake is in step 4. When you multiply source voltage by current, you are calculating the power dissipation of the entire circuit. Instead, you need to calculate the power dissipation of only the heater, i.e., the 24.2 Ω resistor. You can use P = (I^2)R for this. You know the current through the heater, so square the current and multiply it by the heater resistance: I = 110 V/30.2 Ω = 3.642 A; P = (3.642 A)(3.642 A)(24.2 Ω) = 321 W.
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