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17.5: Combined Series-Parallel Circuits

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    Complicated circuit of a polynomial plotter
    Figure 17.5.1

    Electrical circuits can become immensely complicated. This circuit is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various x values.

    Combined Series-Parallel Circuits

    Most circuits are not just a series or parallel circuit; most have resistors in parallel and in series. These circuits are called combination circuits. When solving problems with such circuits, use this series of steps.

    1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them.
    2. For resistors in series, calculate the single equivalent resistance that can replace them.
    3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The voltage drops and currents though individual resistors can then be calculated.

    Example 17.5.1

    In the combination circuit sketched below, find the equivalent resistance for the circuit, find the total current through the circuit, and find the current through each individual resistor.

    Example problem circuit with a pair of parallel resistors in series with another resistor
    Figure 17.5.2


    We start by simplifying the parallel resistors R2 and R3.

    1/R23=(1/180 Ω)+(1/220 Ω)=199 Ω

    R23=99 Ω

    We then simplify R1 and R23 which are series resistors.

    RT=R1+R23=110 Ω+99 Ω=209 Ω

    We can then find the total current, IT=VT/RT=24 V/209 Ω=0.11 A.

    All the current must pass through R1, so I1=0.11 A.

    The voltage drop through R1 is (110 Ω)(0.11 A)=12.6 volts.

    Therefore, the voltage drop through R2 and R3 is 11.4 volts.

    I2=V2/R2=11.4 V/180 Ω=0.063 A and I3=V3/R3=11.4 V/220 Ω=0.052 A

    Use the Marquee Lights simulation below to arrange many identical light bulbs in different configurations. Try to add several bulbs in series and observe the circuit diagram to see what happens to the current, resistance, and brightness of the bulbs. Then, do the same with several bulbs in parallel and compare. Lastly, set the Configuration slider to Mixed and observe what happens in a combined series-parallel circuit:

    Interactive Element


    • Combined circuit problems should be solved in steps.


    1. Two 60.0 Ω resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0 Ω resistor. The combination is placed across a 120. V potential difference.
      1. Draw a diagram of the circuit.
      2. What is the equivalent resistance of the parallel portion of the circuit?
      3. What is the equivalent resistance for the entire circuit?
      4. What is the total current in the circuit?
      5. What is the voltage drop across the 30.0 Ω resistor?
      6. What is the voltage drop across the parallel portion of the circuit?
      7. What is the current through each resistor?
    2. Three 15.0 Ω resistors are connected in parallel and the combination is then connected in series with a 10.0 Ω resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit.

    Explore More

    Use this resource to answer the questions that follow.

    1. In a circuit that contains both series and parallel parts, which parts of the circuit are simplified first?
    2. In the circuit drawn below, which resistors should be simplified first?
    Practice circuit with a total of four resistors and a current source
    Figure 17.5.3

    Additional Resources

    Study Guide: Electrical Systems Study Guide

    Real World Application: Christmas Lights

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