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Circuits: Power used by resistors in parallel and series 2014 Teachable Tidbit Jared Rovny, Evan Finch, Nathan Flowers-Jacobs

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Where this will fit We’re targeting a course like PHYS 181 (2 nd semester physics with calculus) We assume the students have already been exposed to the concepts of current and voltage. They will also have seen the following ideas: Ohm’s Law ( V=IR for resistors) Ideal batteries maintain a constant voltage difference between two terminals Power = Voltage x Current for resistors, this means P=I 2 R and P=V 2 /R for lightbulbs, brightness is proportional to power used Resistors in parallel and series Two resistors in series have the same current through them and thus are equivalent to a single resistor with resistance ( R eq =R 1 +R 2 ) Two resistors in parallel have the same voltage drop across them and thus are equivalent to a single resistor with resistance R eq =1/(1/R 1 +1/R 2 )

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Combinations of Resistors Learning Goals Develop the concepts of current and voltage in circuits. Understand power used by resistors in series or in parallel. Learning Objectives Build and Analyze circuits with 2 resistors (lightbulbs) in parallel or series. Qualitatively understand voltage drops and current flows in a 3-bulb circuit.

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Resistors (lightbulbs) in series These two resistors are in series so: I 1 = I 2 R eq = R 1 +R 2 Power in a resistor P = I 2 R. These resistors have the same current. if R 1 =R 2, then P 1 =P 2, so the two bulbs will be equally bright! R2R2 R1R1

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Resistors (lightbulbs) in parallel These two resistors are in parallel so V 1 =V 2. Power used by a resistor is P=V 2 /R. These resistors have the same voltage drop. If R 1 =R 2, they must be using equal power, so these two bulbs will be equally bright! R1R1 R2R2

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Clicker question All of the bulbs in both circuits are 5 Ω. We know: Bulbs 1 and 2 are equally bright Bulbs 3 and 4 are equally bright. Now compare the top set (parallel) to the bottom set (series). A) The bulbs in parallel will be brighter. B) The bulbs in series will be brighter C) All bulbs will be equally bright since energy from the battery is conserved 9V 1 2 V 3 4

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Clicker question All of the bulbs in both circuits are 5 Ω. We know: Bulbs 1 and 2 are equally bright Bulbs 3 and 4 are equally bright. Now compare the top set (parallel) to the bottom set (series). A) The bulbs in parallel will be brighter. B) The bulbs in series will be brighter C) All bulbs will be equally bright since energy from the battery is conserved 9V 1 2 V 3 4 Using the batteries and bulbs being passed out to you, work in groups to build both circuits and find out!

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The parallel bulbs were brighter-why? How can we find the current through the series bulbs? How can we find the voltage across the parallel bulbs? Can we use both P=V 2 /R and P=I 2 R in either circuit? Power is greater in the parallel circuit-is this consistent with energy conservation?

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Group Activity With the same circuit-building materials: Each group will build a circuit with 3 bulbs (5 ohm resistors) and one battery. Try to make all 3 bulbs come on, but not all having the same brightness (i.e. two bulbs may be equally bright, but the other should be different). Each person should then draw a circuit diagram of the circuit you have built, and start thinking about how it works.

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Wrap up How many different circuit configurations are possible? Here is a diagram for a circuit you might have built. Let’s discuss: which bulbs are the same brightness? Why? HW: Calculate the power (in Watts) used in each bulb! + - 9V

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Series Circuits AKA Voltage Dividers Cardinal Rules for Series Circuits Potential difference is divided proportionately amongst each of the resistors placed.

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