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1/81 Circuit Examples Identify and describe series and parallel circuits. Simplify those circuits. Calculate the current and/or voltage of each part of a circuit. By observing and participating in this lesson, you will be able to:

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2/81 Series and Parallel Circuits Electrical components can be connected in various ways. This drastically changes the properties of the circuit.

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3/81 Series Circuits One simple way to arrange components in an electrical circuit is to create one large continuous loop with the components: 2 batteries switch Light bulb resistor This is similar to a TV series where one episode follows another.

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4/81 Parallel Circuits Another way to connect a circuit is in parallel. In this arrangement, each component is connected separately in its own loop. 2 batteries 3 resistors in parallel

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5/81 Resistors in Parallel To find the equivalent resistance of resistors added in parallel: Parallel or total combined resistance, Ω Individual resistors, Ω

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6/81 Parallel Resistor Example Calculate the total effective resistance of two 10 Ohm resistors connected in parallel. Take the reciprocal of each side of equation Combine, use common denominator if needed

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7/81 Parallel Resistor Observations 5 Ohms. Notice that the total overall resistance is lower than either one of them individually! This occurs because there are multiple paths for the electrons to take, lowering their resistance.

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8/81 Circuit Problems This section will detail how to calculate the various electrical quantities in a circuit.

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9/81 Series Circuit Calculate the current and electric potential difference for each component of the circuit shown. R 2 =10Ω R 1 =5Ω 5V A good first step is to simplify the circuit.

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10/81 Circuit Simplification R 2 =10Ω R 1 =5Ω 5V Because this is a series circuit, to combine the resistors and simplify the circuit, they are merely added together. 5V Rs=15Ω

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11/81 Current in a Series Circuit 5V Rs=15Ω Use the voltage of the power supply and the total resistance of the circuit to find the total current flowing through the circuit. Because the electron flow has no where else to go, this amount is also the current flowing through both resistors. I 1 and I 2 is that same.33 Amperes.

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12/81 Voltage in a Series Circuit R 2 =10Ω R 1 =5Ω 5V Since we know the current flowing through each resistor, we can use Ohms law to find the potential difference for each of those resistors. Notice how the sum of the two voltages adds up to the power supply for the circuit.

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13/81 Parallel Circuit Calculate the current and electric potential difference for each component of the circuit shown. 5V R 1 =5ΩR 2 =10Ω Notice how this parallel circuit contains the exact same components as the series circuit, they are just arranged differently. Again, a good first step is to simplify the circuit.

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14/81 Circuit Simplification 5V R 1 =5ΩR 2 =10Ω Notice this parallel resistance is less than either one individually.

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15/81 Shortcut Formula An equivalent formula can be used for two resistors, R 1 and R 2, connected in parallel. Sometimes this formula is easier to manipulate. It may be easier to remember this formula as the product over the sum for the two resistors.

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16/81 Voltage in a Parallel Circuit 5V R 1 =5ΩR 2 =10Ω The easy part about any parallel circuit is the voltage applied to each item. Since each item has its own independent connection to the battery or power supply, each item receives that potential. In this case, V 1 and V 2 are each 5V.

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17/81 Current in a Parallel Circuit 5V R 1 =5ΩR 2 =10Ω Once you realize that the electric potential for each resistor is 5V, finding the current is easy using Ohms law, V=IR.

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18/81 Current Observations 5V R 1 =5ΩR 2 =10Ω Notice the two currents add up to the same value as the total current in the circuit. This is a good way to check your work.

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19/81 Conclusion You should now be able to describe series and parallel circuits with words and numbers! Questions??? Homework: Check with your instructor

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