Download presentation

Presentation is loading. Please wait.

Published byMitchel Paxson Modified over 2 years ago

1
Series Circuits: Other examples:

2
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

3
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

5
Ohm’s Law V = IR I = V/R

6
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

7
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

8
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 + -

9
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

10
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

11
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

12
__________ 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.

13
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

Similar presentations

OK

DC Circuits: Review Current: The rate of flow of electric charge past a point in a circuit Measured in amperes (A) 1 A = 1 C/s = 6.25 1018 electrons.

DC Circuits: Review Current: The rate of flow of electric charge past a point in a circuit Measured in amperes (A) 1 A = 1 C/s = 6.25 1018 electrons.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on regional trade agreements signed Ppt on factoring polynomials Ppt on panel discussion questions Ppt on mars one way Ppt on money and credit download Ppt on reuse of wastewater Ppt on trial and error theory Ppt on micro operations Ppt on db2 mainframes definition Ppt on nitrogen cycle and nitrogen fixation lightning