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Series Circuits: Other examples:

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Series circuits - ________________________________________ _________________________________________ Assume: 1. _____________________________________________________ 2. _____________________________________________________ 3. _____________________________________________________ 4. _____________________________________________________ Wires have no potential drop (voltage) across them Pos. current is out of the “high” voltage side circuit element ___ voltage source circuit element ___ circuit element ___ wire have only 1 path for current + ________ potential ________ potential - wire All energy provided by source is used in the elements No charge is “lost.” All current returns to the source. high low I I 1 2 could have switches, etc. 3

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V V1V1 V2V2 I I1I1 R1R1 R2R2 I2I2 _______________ Conservation: V = _______________ Conservation: I = _______________ (Total) Resistance: R eq = __________ Law applies to the total : V = and to each individual element: V 1 = V 2 = For a circuit with 2 resistors: Energy V 1 + V 2 Charge I 1 = I 2 Equivalent R 1 + R 2 Ohm’s I R eq I1R1I1R1 I2R2I2R2

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Ohm’s Law V = IR I = V/R

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Ex. Find all the voltages and currents in the circuit below: 20. V 40 120 V 1 = I 1 = R 1 = V 2 = I 2 = R 2 = V = I = R eq = 20. V 40 120 160 20/160 A 0.125 A 5.0 V 15.0 V 0.125 A R eq = R 1 + R 2 V 1 = I 1 R 1 I = V/R eq I = I 1 = I 2 V 2 = I 2 R 2

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V = 20. V 40 120 V “divides up” ______________________________ as the R’s This is because ___________ R requires _________ energy. Series circuits are _______________________________. in the same proportion voltage dividers Form the _________ of each resistance to R eq = _______, and then multiply by the __________ voltage V 40 160 x 20 V = 5 V 120 160 x 20 V = 15 V 160 R1R1 R eq R2R2 ratio total more

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Plot V vs. “distance around circuit.” 20 15 wire V dropped across the ______ resistor wire 40 wire 0 back to ____ side of the battery ______ drop across wires because we assume _________ distance around circuit No R = 0 potential difference (V) ____ side of battery + V dropped across the ______ R 160 at the ___ side of the battery + -

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Important : “I is ______________ everywhere in ___________ circuit” does NOT mean that I is ___________ in _________________ circuit! 10. V R1=R1= R 2 = I= I 1 = I 2 = 10. V R2=R2= R3=R3= I= I 1 = I 2 = I 3 = I is still the _______________ in all parts of the second circuit, but it is a ________________ I than the first one! the same a series the same every other 10/100 =.1 A.1 A 10/200 =.05 A.05 A same different R1=R1= 75 25 100 25 75 .05 A

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Equivalent resistance: _________________________________ ________________________________________________________ The total I = 20. V 40 120 If you replace the resistors of a circuit with one resistor, the total I would be the same Replacing this part of the circuit with a single _______________ resistor: R eq = R 1 + R 2 = = 40 + 120 equivalent 160 …gives you this circuit: R eq = 160 20. V V R eq = 20 V 160 = 0.125 A This is the ____________ as before. same I

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50 150 V = 20 V 70 20 16 8 8 V = 12 V 10 20 All _______________ circuits can be ___________________ in this way. This can be done even if the ______________________ is not shown. simplified series voltage source 15 5 5 R eq = _____ V = 20 V V = 12 V R eq results in the _____________ as the _________________ circuit. A. B. C. D. 200 R eq = _____ 24 R eq = _____ 90 R eq = _____ 50 same I original

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__________ Hookups: V R1R1 R2R2 Original circuit: To measure I 1, the current through R 1, _________________ the circuit and _____________ an ________________ next to R 1 Meter ammeter disconnect insert V R1R1 R2R2 V R1R1 R2R2 A Other possibilities: V R1R1 R2R2 V R1R1 R2R2 A A ___ is the same everywhere, so _________________________I disconnect insert anywhere gives same I.

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To measure, V 1, the voltage across R 1, __________disconnect the circuit. Simply connect the ______________ across R 1 voltmeter do NOT Other possibilities: V R1R1 R2R2 V R1R1 R2R2 V Original circuit: V R1R1 R2R2 V V R1R1 R2R2 V Similarly, to measure the _________ voltage V or V 2 : V R1R1 R2R2 V V total V R1R1 R2R2

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