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# Aut 241 Auto Electricity and Electronics - PowerPoint PPT Presentation

Aut 241 Auto Electricity and Electronics. Series – Parallel Circuits Chapter 7. SERIES-PARALLEL CIRCUITS. Series-parallel circuits are a combination of series and parallel segments in one complex circuit.

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Aut 241Auto Electricity and Electronics

• Series – Parallel Circuits

• Chapter 7

• Series-parallel circuits are a combination of series and parallel segments in one complex circuit.

• A series-parallel circuit includes both parallel loads or resistances, plus additional loads or resistances that are electrically connected in series.

• A circuit where the load is in series with other loads is parallel.

FIGURE 7-1 A series-parallel circuit.

• A circuit where a parallel circuit contains resistors or loads are in series in one or more branches.

This complete headlight circuit with all bulbs and switches is a series-parallel circuit.

SERIES-PARALLEL CIRCUITSSeries-Parallel Circuit Faults

• If a conventional parallel circuit, such as a taillight circuit, had an electrical fault that increased the resistance in one branch of the circuit, then the amount of current flow through that one branch will be reduced.

• The key to solving series-parallel circuit problems is to combine or simplify as much as possible.

Solving a series-parallel circuit problem.

SERIES-PARALLEL CIRCUITEXAMPLES

• Each of the four examples includes solving for the following:

• Total resistance

• Current flow (amperes) through each branch, as well as total current flow

• Voltage drop across each resistance

• A series-parallel circuit is called a compound circuit or a combination circuit.

• A series-parallel circuit is a combination of a series and a parallel circuit, which does not include fuses or switches.

• A fault in a series portion of the circuit would affect the operation if the series part was in the power or the ground side of the parallel portion of the circuit.

• A fault in one leg of a series-parallel circuit will affect just the component(s) in that one leg.

• Explain why an increase in resistance in the series part of a series-parallel circuit will affect the current (amperes) through the parallel legs (branches).

• What would be the effect of an open circuit in one leg of a parallel portion of a series-parallel circuit?

• What would be the effect of an open circuit in a series portion of a series-parallel circuit?

• Half of the dash is dark. Technician A says that a defective dash light dimmer can be the cause because it is in series with the bulbs that are in parallel. Technician B says that one or more bulbs could be defective. Which technician is correct?

• Technician A only

• Technician B only

• Both Technicians A and B

• Neither Technician A nor B

2. All brake lights are dimmer than normal. Technician A says that bad bulbs could be the cause. Technician B says that high resistance in the brake switch could be the cause. Which technician is correct?

• Technician A only

• Technician B only

• Both Technicians A and B

• Neither Technician A nor B

3. See Figure 7-8 to solve for total resistance (RT) and total current (IT).

• 10 ohms ÷ 1.2 A

• 4 ohms ÷ 3 A

• 6 ohms ÷ 2 A

• 2 ohms ÷ 6 A

4. See Figure 7-9 to solve for the value of R3and total resistance (RT).

• 12 ohms ÷ 12 ohms

• 1 ohm ÷ 7 ohms

• 2 ohms ÷ 8 ohms

• 6 ohms ÷ 6 ohms

5. See Figure 7-10 to solve for voltage (E ) and total resistance (RT).

• 16.3 volts ÷ 12 ohms

• 3.3 volts ÷ 2.4 ohms

• 1.36 volts ÷ 1 ohm

• 6 volts ÷ 4.4 ohms

6. See Figure 7-11 to solve for R1 and total resistance (RT).

• 3 ohms ÷ 15 ohms

• 1 ohm ÷ 15 ohms

• 2 ohms ÷ 5 ohms

• 5 ohms ÷ 5 ohms

7. See Figure 7-12 to solve for total resistance (RT) and total current (I ).

• 3.1 ohms ÷ 7.7 amperes

• 5.1 ohms ÷ 4.7 amperes

• 20 ohms ÷ 1.2 amperes

• 6 ohms ÷ 4 amperes

8. See Figure 7-13 to solve for the value of E and total resistance (RT).

• 13.2 volts ÷ 40 ohms

• 11.2 volts ÷ 34 ohms

• 8 volts ÷ 24.2 ohms

• 8.6 volts ÷ 26 ohms

9. See Figure 7-14 to solve for total resistance (RT) and total current (I ).

• 1.5 ohms ÷ 8 amperes

• 18 ohms ÷ 0.66 ampere

• 6 ohms ÷ 2 amperes

• 5.5 ohms ÷ 2.2 amperes

10. See Figure 7-15 to solve for total resistance (RT) and total current (I ).

• 48 ohms ÷ .42 ampere

• 20 ohms ÷ 1 ampere

• 30 ohms ÷ .66 ampere

• 10.2 ohms ÷ 1.96 amperes