All About Circuits

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).

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  • Question 11 of 27

    Re-draw this circuit in the form of a schematic diagram:




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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.

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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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