a b  R C I I R  R I I r V

Yesterday Ohm’s Law V=IR Ohm’s law isn’t a true law but a good approximation for typical electrical circuit materials Resistivity  =1/  (Conductivity): Property of the material Resistance proportional to resistivity and length, inversely proportional to area

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. Compare the resistance of the two cylinders. a) R 1 > R 2 b) R 1 = R 2 c) R 1 < R 2 Question 1

1.a 2.b 3.c

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. Compare the resistance of the two cylinders. a) R 1 > R 2 b) R 1 = R 2 c) R 1 < R 2 Question 1 Resistance is proportional to Length/Area

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors: a) v 1 > v 2 b) v 1 = v 2 c) v 1 < v 2 Question 2

1.a 2.b 3.c

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors: a) v 1 > v 2 b) v 1 = v 2 c) v 1 < v 2 Question 2 Current  Area  Current Density Current Density  average velocity of electrons I is the same A 1 v 2

By Ohm’s law, the Voltage difference across resistance R 1 is Across R 2 is Total voltage difference  the effective single resistance is Resistors in Series a c R effective a b c R1R1 R2R2 I What is the same effective single resistance to two resistances in series? Whenever devices are in SERIES, the current is the same through both.

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...

The World’s Simplest (and most useful) circuit: Voltage Divider By varying R 2 we can controllably adjust the output voltage! V0V0 R1R1 R2R2 V

Two resistors are connected in series to a battery with emf E. The resistances are such that R 1 = 2R 2. The currents through the resistors are I 1 and I 2 and the potential differences across the resistors V 1 and V 2. Are: Question 3 a)I 1 >I 2 and V 2 =E b)I 1 =I 2 and V 2 = E c)I 1 =I 2 and V 2 =1/3E d)I 1 <I 2 and V 2 =1/2E e)I 1 <I 2 and V 2 =1/3E

Resistors in ParallelParallel a d I I R1R1 R2R2 I1I1 I2I2 V I a d I RV Current through R 1 is I 1. Current through R 2 is I 2. Very generally, devices in parallel have the same voltage drop But current is conserved  

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

Kirchhoff’s First Rule “Loop Rule” or “Kirchhoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." KVL: This is just a restatement of what you already know: that the potential difference is independent of path!  R1R1  R2R2 I     IR 1  IR 2     0 0

Rules of the Road  R1R1  R2R2 I     IR 1  IR 2    Our convention: Voltage gains enter with a + sign, and voltage drops enter with a  sign. We choose a direction for the current and move around the circuit in that direction. When a battery is traversed from the negative terminal to the positive terminal, the voltage increases, and hence the battery voltage enters KVL with a + sign. When moving across a resistor, the voltage drops, and hence enters KVL with a  sign. 

Current in a Loop b a d e c f  R1R1 I R2R2 R3R3 R4R4 I    KVL: Start at point a (could be anywhere) and assume current is in direction shown (could be either)

Question 3 (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –Just 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 ? R 12V R I a b

Question 3 1.a 2.b 3.c

Question 3 (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 From symmetry the potential ( V a - V b ) before the switch is closed is V a - V b = +12V. Therefore, when the switch is closed, potential stays the same and NO additional current will flow! Therefore, the current before the switch is closed is equal to the current after the switch is closed. Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –Just 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 ? R 12V R I a b

Question 3 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 Write a loop law for original loop: 12V  I 1 R = 0 I 1 = 12V/R Write a loop law for the new loop: 12V +12V  I 0 R  I 0 R = 0 I 0 = 12V/R R 12V R I a b

Kirchhoff’s Second Rule “Junction Rule” or “Kirchhoff’s Current Law (KCL)” In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff'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. The currents entering and leaving circuit nodes are known as “branch currents”. Each distinct branch must have a current, I i  assigned to it

How to use Kirchhoff’s Laws A two loop example: Assume currents in each section of the circuit, identify all circuit nodes and use KCL.  1  I 1 R 1  I 2 R 2 = 0 (3)I 2 R 2   2  I 3 R 3 = 0    I 1 R 1    I 3 R 3   1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 (1) I 1 = I 2 + I 3 Identify all independent loops and use KVL.

How to use Kirchoff’s Laws  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 Solve the equations for I 1, I 2, and I 3 : First find I 2 and I 3 in terms of I 1 :  From eqn. (2) From eqn. (3) Now solve for I 1 using eqn. (1):

Let’s plug in some numbers  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3  1 = 24 V  2 = 12 V R 1 = 5  R 2 =3  R 3 =4  Then, and I 1 =2.809 A I 2 = A, I 3 = A

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

Summary Kirchhoff’s Laws –KCL: Junction Rule (Charge is conserved) –Review KVL ( V is independent of path) Non-ideal Batteries & Power Discharging of capacitor through a Resistor: Reading Assignment: Chapter 26.6 Examples: 26.17,18 and 19

Two identical light bulbs are represented by the resistors R 2 and R 3 (R 2 = R 3 ). The switch S is initially open. 2) If switch S is closed, what happens to the brightness of the bulb R 2 ? a) It increases b) It decreases c) It doesn’t change 3) What happens to the current I, after the switch is closed ? a) I after = 1/2 I before b) I after = I before c) I after = 2 I before

I E R1R1 R4R4 R2R2 R3R3 Four identical resistors are connected to a battery as shown in the figure. 5) How does the current through the battery change after the switch is closed ? a) I after > I before b) I after = I before c) I after < I before Before: R tot = 3R I before = 1/3 E/R After: R 23 = 2R R 423 = 2/3 R R tot = 5/3 R I after = 3/5 E/R