Review Lecture 1

Copyright © 2009 Pearson Education, Inc. Admin for next week: Mid-term date Friday March 20 th. Labs as usual NO DISCUSSION SECTIONS THIS WEEK Extra Office hours 2-4pm Monday: Dr Holder Sharp Lab pm Tuesday: Dr Holder Sharp Lab pm Wednesday: Seth Sharp Lab pm Thursday: Zach Sharp Lab 320 is also OK – or stop by my office any time Monday or Tuesday I will be away Wednesday – Friday. Dr Ji will give Wednesday’s class and the exam.

Copyright © 2009 Pearson Education, Inc. Admin: Assignments : Fifth Assignment is posted. Not due until Monday (23 rd ), but good exam practice Three questions on Thevenin circuits – I’ll go through one of them today

Copyright © 2009 Pearson Education, Inc. What does the test cover? What are current, voltage and charge? Ohm’s law: V=IR Resistivity and resistance Temperature dependence What is power? P=IV Resistors in series and parallel Calculating equivalent resistance The voltage divider rule The current divider rule Ammeters, Voltmeters and Ohmmeters Voltage sources and current sources Ideal and Practical sources Combining sources

Copyright © 2009 Pearson Education, Inc. What does the test cover? Kirchhoff’s rules Junction rule (current law) Loop rule (voltage law) Circuit analysis Applying Kirchhoff’s rules Mesh analysis Thevenin/Norton Equivalents

Copyright © 2009 Pearson Education, Inc. What formulae will I have? Note that the equations for current, voltage, power, etc, are not given, because they are implicit in the definition: if you don’t know the equation, you don’t understand what they are. Any required constants will be given (e.g. resistivity values)

Copyright © 2009 Pearson Education, Inc. What will be the format of the exam? Roughly… 5 short answer questions: 2 points each. 4 longer questions: 5 points each. Rules Closed book exam – no notes or textbooks. Please seat yourselves with at least one space between you. A scientific calculator is required Graphing calculators are OK – but I can’t give partial credit for the algebra if you don’t show it. No phones/ phone calculators Students with accommodations can use the test center – but fill out the DSS form.

Copyright © 2009 Pearson Education, Inc. How should I study?

Copyright © 2009 Pearson Education, Inc. How should I study? Hard!

Copyright © 2009 Pearson Education, Inc. How should I study? Take it seriously: 15% of your grade I aim for “fair but challenging” The best resource is the lecture notes – in particular, make sure you can solve all of the worked examples, without the solutions in front of you. The textbook (Ch 25 & 26) is also useful. Review the discussion problems, and homework problems. Come to office hours (soon!) if anything is not clear – or drop me an , but office hours are usually much more effective

Copyright © 2009 Pearson Education, Inc. Clearing up a misconception. Current flows through the elements of a circuit. Voltage is measured across the elements of a circuit. Current only changes if there is a junction. The voltage between two points is the same, regardless of the path taken to calculate it.

Copyright © 2009 Pearson Education, Inc. Light bulbs Power defines how bright a light bulb gets. P=IV, P=I 2 R, P=V 2 /R Rank these light bulbs in brightness, each with the same resistance: What happens to A if we unscrew E? Total resistance increases Current decreases (V=IR) Bulb A gets dimmer! 1. C 2. A & B 3. D & E

Copyright © 2009 Pearson Education, Inc. Light bulbs Power defines how bright a light bulb gets. P=IV, P=I 2 R, P=V 2 /R Two lightbulbs of resistance R1 and R2 (R2 > R1) and a battery are all connected in series. (a) Which bulb is brighter? (b) What if they are connected in parallel? Draw it! A: the bulb with resistance R2 B: the bulb with resistance R1

Copyright © 2009 Pearson Education, Inc. Voltage dividers What is the voltage across the 2 Ohm resistor? Use intuition… remember Kirchhoff.. The voltage across each R is split according to the fraction of the total resistance V R = V IN (R/R EQ )

Copyright © 2009 Pearson Education, Inc. Current dividers What is the current through resistor R2? Use intuition… The current through each R is split according to the inverse of the fraction of the total resistance. I R = I IN (R EQ /R) (note that R EQ is always less than R – so I R cannot be greater than I IN ) 3 Amps 10Ω 20Ω

Copyright © 2009 Pearson Education, Inc. Thevenin Equivalent (a freebie)

Copyright © 2009 Pearson Education, Inc.

Mesh analysis VS1=12V VS2 =5V R1 =50Ω R2 =R3 =20Ω R4 =10Ω R5 =15Ω Find the mesh currents, and the voltage across R4

Copyright © 2009 Pearson Education, Inc. Hint - you can use this to check your algebra:

Copyright © 2009 Pearson Education, Inc. Kirchhoff’s law circuit analysis Example 26-10: Jump starting a car. A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and internal resistance Ω. Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 Ω. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor can be represented as a resistor R s = 0.15 Ω. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected.

Copyright © 2009 Pearson Education, Inc. See page 687 for full solution I 1 =76A I 2 =-5A (charging) I 3 =71A Kirchhoff’s law circuit analysis