All About Circuits

DC Electric Circuits

Series-Parallel DC Circuits


41 questions By Tony R. Kuphaldt

Page 12 of 14 0 of 41 answers revealed (0%)
  • Question 34 of 41

    A student built this resistor circuit on a solderless breadboard, but made a mistake positioning resistor R3. It should be located one hole to the left instead of where it is right now:



    Determine what the voltage drop will be across each resistor, in this faulty configuration, assuming that the battery outputs 9 volts.

    • R1 = 2 k Ω VR1 =
    • R2 = 1 k Ω VR2 =
    • R3 = 3.3 k Ω VR3 =
    • R4 = 4.7 k Ω VR4 =
    • R5 = 4.7 k Ω VR5 =
    Reveal answer
  • Question 35 of 41

    Suppose you were designing a circuit that required two LEDs for “power on” indication. The power supply voltage is 15 volts, and each LED is rated at 1.6 volts and 20 mA. Calculate the dropping resistor sizes and power ratings:



    After doing this, a co-worker looks at your circuit and suggests a modification. Why not use a single dropping resistor for both LEDs, economizing the number of components necessary?



    Re-calculate the dropping resistor ratings (resistance and power) for the new design.

    Reveal answer
  • Question 36 of 41

    Calculate all voltages and currents in this circuit:



    The battery voltage is 15 volts, and the resistor values are as follows:

    R1 = 1 kΩ
    R2 = 3.3 kΩ
    R3 = 4.7 kΩ
    R4 = 2.5 kΩ
    R5 = 10 kΩ
    R6 = 1.5 kΩ
    R7 = 500 Ω
    Reveal answer

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  • Alana Ben May 19, 2020

    How did you guys find the answer for Q5, figure 6

    Like. Reply
    • The Mullet May 26, 2020
      I don't know if there is an easier way, but I was able to calculate the resistance using the delta-y conversion for a bridge resistor circuit. https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/
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    • RK37 May 28, 2020
      The 330 Ω and 470 Ω resistors are in parallel; they don't look like they're in parallel, but note that their terminals are connected to the same two nodes. REQ for those two resistors (call it REQ1) is ~193.9 Ω. If you redraw the circuit with REQ1, there are two (parallel) current paths between node A and node B: one consisting of the 220 Ω resistor and REQ1 in series, and the other consisting of the 100 Ω resistor. REQ2 = REQ1 + 220 Ω = 413.9 Ω, and then RAB = 100 Ω || REQ2 = 80.54 Ω.
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