Ch 26 Circuits_2009

# Ch 26 Circuits_2009 - Go over mid term first Direct Current...

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Direct Current Circuits Chapter 26 In this chapter we discuss only direct currents used in such circuits as flashlights and automobile circuits . Household appliances use alternating currents and are not discussed in this section. Go over mid term first

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One of the most common direct current circuit elements [and alternating current] is the resistor.

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Resistors can occur in combinations with other resistors. We will study resistors that occur in series and parallel circuits as shown in the diagrams to the right. The question is what then will be the equivalent resistance of resistors in parallel and series combinations ? That is, if you replace any of the network circuits to the right by a single resistor, how would we determine that single resistance? Such that: I V R ab eq =
Resistors in Series If the resistors are in series the current must be the same through all of them ( moving charge is not used up as it passes through a circuit ). Applying Ohm’s law we have for the series circuit: V ax = IR 1 ; V xy = IR 2 ; V yb = IR 3 Furthermore, since potential is related to work and work adds, thus electrical potential adds due to each resistor, which takes Joule energy away from the electrons.

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The potential difference across each resistor in general will not be the same. Nonetheless, potential like work adds hence: V ab = V ax + V xy + V yb = IR 1 + IR 2 + IR 3 V ab = I(R 1 + R 2 + R 3 ) Thus And so 3 2 1 R R R I V R ab eq + + = = Simply put resistors in series add.
Resistors in Parallel Like water through a pinched versus an open hose, resistors in parallel can have different amounts of flow of electrons (current) through them. However, just like “ water pressure ”, resistors like hoses connected in parallel will have the same pressure ; that is the same electric potential or voltage drop across each circuit element or resistor R 1 , R 2 and R 3 .

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2 2 R V I ab = 1 1 R V I ab = 3 3 R V I ab = The total current in is simply the sum of the currents through each resistor. + + = + + = + + = 3 2 1 3 2 1 3 2 1 1 1 1 R R R V R V R V R V I I I I ab ab ab ab That is: eq ab R R R R V I 1 1 1 1 3 2 1 = + + =
And so we see for resistors in parallel the equivalent resistance is given by: + + + + = n eq R R R R R 1 ... 1 1 1 1 3 2 1 The equivalent resistance is always less than any individual resistance.

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Compute the equivalent resistance of the network from the figure below and find the current in each resistor:
To find the equivalent resistance we start with the 6 Ohm and the 3 Ohm resistors in parallel. Hence we use the parallel resistor formula. + + + + = n eq R R R R R 1 ... 1 1 1 1 3 2 1 2 1 6 3 3 1 6 1 1 = = + = eq R Hence the total parallel resistance is R eq = 2Ω

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Now we have two resistors in series and so we simply add these two resistors to get the equivalent resistance.
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