Physics 1202: Lecture 8 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #3:Homework #3: –On Masterphysics today: due Friday next week –Go to masteringphysics.com

Today’s Topic : Electric current (Chap.17) Review of –Electromotive force (battery) –Power Circuits –Devices –Resistance in series & in parallel –Kirchorff’s rules –RC circuits

Power Batteries & Resistors Energy expended What’s happening? Assert: chemical to electrical to heat Charges per time Energy “drop” per charge Units okay? For Resistors: Rate is:

Batteries (non-ideal) Parameterized with “internal resistance” r in series with   : “electromotive force” (emf)  R I I r V Power delivered to the resistor R: P max when R/r =1 ! = V( I=0 )  - Ir - IR = 0  - Ir = V 

 R I  = R I

Devices Conductors: Purpose is to provide zero potential difference between 2 points. »Electric field is never exactly zero.. All conductors have some resistivity. »In ordinary circuits the conductors are chosen so that their resistance is negligible. Batteries (Voltage sources, seats of emf): Purpose is to provide a constant potential difference between 2 points. »Cannot calculate the potential difference from first principles.. electrical  chemical energy conversion. Non-ideal batteries will be dealt with in terms of an "internal resistance". +- V + - OR

Devices Resistors: Purpose is to limit current drawn in a circuit. »Resistance can be calculated from knowledge of the geometry of the resistor AND the “resistivity” of the material out of which it is made. »The effective resistance of series and parallel combinations of resistors will be calculated using the concepts of potential difference and current conservation (Kirchoff’s Laws). Resistance Resistance is defined to be the ratio of the applied voltage to the current passing through. V I I R UNIT: OHM = 

Resistors in Series The Voltage “drops”: Whenever devices are in SERIES, the current is the same through both ! This reduces the circuit to: a c R effective a b c R1R1 R2R2 I Hence:

Another (intuitive) way... Consider two cylindrical resistors with lengths L 1 and L 2 V R1R1 R2R2 L2L2 L1L1 Put them together, end to end to make a longer one...

Resistors in Parallel What to do? But current through R 1 is not I ! Call it I 1. Similarly, R 2  I 2. How is I related to I 1 & I 2 ?? Current is conserved! a d a d I I I I R1R1 R2R2 I1I1 I2I2 R V V Very generally, devices in parallel have the same voltage drop  

Another (intuitive) way... Consider two cylindrical resistors with cross-sectional areas A 1 and A 2 Put them together, side by side … to make a “fatter” one with A=A 1 +A 2, V R1R1 R2R2 A1A1 A2A2 

V R1R1 R2R2 V R1R1 R2R2 Summary Resistors in series –the current is the same in both R 1 and R 2 –the voltage drops add Resistors in parallel –the voltage drop is the same in both R 1 and R 2 –the currents add

Lecture 8, ACT 1 I have two identical light bulbs. First I hook them up in series. Then I hook them up in parallel. In which case are the bulbs brighter? (The resistors represent light bulbs whose brightness is proportional to P = I 2 R through the resistor.) V R R A) SeriesB) ParallelC) The same R R

 I1I1   R R R I2I2 I3I3

Kirchoff's First Rule "Loop Rule" or “Kirchoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." This is just a restatement of what you already know: that the potential difference is independent of path! We will follow the convention that voltage gains enter with a + sign and voltage drops enter with a - sign in this equation. RULES OF THE ROAD:  R1R1  R2R2 I Move clockwise around circuit:  R1R1 R2R2  I   IR 1  IR 2     0 0 KVL:

Loop Example a d b e c f  R1R1 I R2R2 R3R3 R4R4 I  KVL:  

Lecture 8, ACT 2 Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –After the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 R 12 V R I

Kirchoff's Second Rule "Junction Rule" or “Kirchoff’s Current Law (KCL)” In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchoff's Second Rule (the junction rule). "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." This is just a statement of the conservation of charge at any given node.

Junction Example Outside loop:Top loop:Junction:  I1I1   R R R I2I2 I3I3

Chapter 8, ACT 3 Consider the circuit shown: –What is the relation between V a -V d and V a -V c ? (a) (V a -V d ) < (V a -V c ) (b) (V a -V d ) = (V a -V c ) (c) (V a -V d ) > (V a -V c ) 2A (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 2B – What is the relation between I 1 and I 2 ? 12 V I1I1 I2I2 a b d c 50  20  80 

Chapter 8, ACT 4 An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is A) I B) I/2 C) I/4 D) I/8 E) 0

RC Circuits Consider the circuit shown: –What will happen when we close the switch ? –Add the voltage drops going around the circuit, starting at point a. IR + Q/C – V = 0 –In this case neither I nor Q are known or constant. But they are related, V a b c R C This is a simple, linear differential equation.

RC Circuits Case 1: Charging Q 1 = 0, Q 2 = Q and t 1 = 0, t 2 = t To get Current, I = dQ/dt Q tt I V a b c R C

RC Circuits c Case 2: Discharging: Q 1 = Q 0, Q 2 = Q and t 1 = 0, t 2 = t To discharge the capacitor we have to take the battery out of the circuit (V=0) To get Current, I = dQ/dt t I Q t V a b c R C

Chapter 8, ACT 5 c Consider the simple circuit shown here. Initially the switch is open and the capacitor is charged to a potential V O. Immediately after the switch is closed, what is the current ? A) I = V O /RB) I = 0 C) I = RCD) I = V O /R exp(-1/RC) We remember the equation for a discharging capacitor, We note that Q max /C = V O At t = 0, exp (0) = 1, so I = V O /R V a b c R C

Electrical Instruments The Ammeter The device that measures current is called an ammeter. Ideally, an ammeter should have zero resistance so that the measured current is not altered. A  I R1R1 R2R2 + -

Electrical Instruments The Voltmeter The device that measures potential difference is called a voltmeter. An ideal voltmeter should have infinite resistance so that no current passes through it. V  R1R1 R2R2 I IvIv I2I2

Problem Solution Method: Five Steps: 1)Focus on the Problem - draw a picture – what are we asking for? 2)Describe the physics -what physics ideas are applicable -what are the relevant variables known and unknown 3)Plan the solution -what are the relevant physics equations 4)Execute the plan -solve in terms of variables -solve in terms of numbers 5)Evaluate the answer -are the dimensions and units correct? -do the numbers make sense?

Example: Power in Resistive Electric Circuits A circuit consists of a 12 V battery with internal resistance of 2  connected to a resistance of 10 . The current in the resistor is I, and the voltage across it is V. The voltmeter and the ammeter can be considered ideal; that is, their resistances are infinity and zero, respectively. What is the current I and voltage V measured by those two instruments ? What is the power dissipated by the battery ? By the resistance ? What is the total power dissipated in the circuit ? Comment on these various powers.

Step 1: Focus on the problem Drawing with relevant parameters –Voltmeter can be put a two places  R I I r A What is the question ? –What is I ? –What is V ? –What is P battery ? –What is P R ? –What is P total ? –Comment on the various P’s V V 10  2  12 V

Step 2: describe the physics What concepts are relevant ? –Potential difference in a loop is zero –Energy is dissipated by resistance What are the known and unknown quantities ? –Known: R = 10 ,r = 2    = 12 V –Unknown: I, V, P’s

Step 3: plan the solution What are the relevant physics equations ? Kirchoff’s first law: Power dissipated: For a resistance

Step 4: solve with symbols Find I:  - Ir - IR = 0  R I I r A Find V: Find the P’s:

Step 4: solve numerically Putting in the numbers

Step 5: Evaluate the answers Are units OK ? – [ I ] = Amperes – [ V ] = Volts – [ P ] = Watts Do they make sense ? – the values are not too big, not too small … – total power is larger than power dissipated in R »Normal: battery is not ideal: it dissipates energy