# a b  R C I I R  R I I r V Lecture 10, ACT 1 Consider the circuit shown: –What is the relation between V a - V d and V a - V c ? (a) (V a -V d ) < (V.

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a b  R C I I R  R I I r V

Lecture 10, ACT 1 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 ) 12V I1I1 I2I2 a b d c 50  20  80  (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B 1A

Lecture 10, ACT 1 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 ) 12V I1I1 I2I2 a b d c 50  20  80  1A Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! ( V a - V d ) = ( V a - V c ) Point d and c are the same, electrically

Lecture 10, ACT 1 (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 1B – What is the relation between I 1 and I 2 ? 1B 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 ) 12V I1I1 I2I2 a b d c 50  20  80  1A Note that: V b -V d = V b -V c Therefore,

Summary of Simple Circuits Resistors in series: Resistors in parallel: Current thru is same; Voltage drop across is IR i Voltage drop across is same; Current thru is V / R i Kirchhoff’s laws: (Tipler problems on Kirchhoff’’s rules)

Batteries (“Nonideal” = cannot output arbitrary current)  R I I r V Parameterized with "internal resistance" 

Internal Resistance Demo As # bulbs increases, what happens to “ R ”??  R I r V How big is “r”?

Power Batteries & Resistors Energy expended chemical to electrical to heat What’s happening? Assert: Rate is: Charges per time Potential difference per charge Units okay? For Resistors: or you can write it as

More complex now…  R C I I Let’s try to add a Capacitor to our simple circuit Recall voltage “drop” on C ? Write KVL: KVL gives Differential Equation ! What’s wrong here? Consider that and substitute. Now eqn has only “ Q ”.

The Big Idea Previously: –Analysis of multi-loop circuits with batteries and resistors. –Main Feature: Currents are attained instantaneously and do not vary with time!! Now: –Just added a capacitor to the circuit. –What changes?? KVL yields a differential equation with a term proportional to Q and a term proportional to I = d Q /dt.

The Big Idea Physically, what’s happening is that the final charge cannot be placed on a capacitor instantly. Initially, the voltage drop across an uncharged capacitor = 0 because the charge on it is zero ! ( V = Q / C ) As current starts to flow, charge builds up on the capacitor, the voltage drop is proportional to this charge and increases; it then becomes more difficult to add more charge so the current slows

Lecture 10, ACT 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ =  /2 R (c) I 0+ = 2  / R a b  R C I I R 2A (a) I  = 0 (b) I  =  /2 R (c) I  > 2  / R – What is the value of the current I  after a very long time? 2B

Lecture 10, ACT 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ =  /2 R (c) I 0+ = 2  / R a b  R C I I R 2A Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+,   IR  0  IR = 0  I =  /2 R

Lecture 10, ACT 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? 2A (a) I 0+ = 0 (b) I 0+ =  /2 R (c) I 0+ = 2  / R (a) I  = 0 (b) I  =  /2 R (c) I  > 2  / R – What is the value of the current I  after a very long time? 2B The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. The voltage across the capacitor is limited to  ; the current goes to 0. a b  R C I I R

Behavior of Capacitors Charging –Initially, the capacitor behaves like a wire. –After a long time, the capacitor behaves like an open switch. Discharging –Initially, the capacitor behaves like a battery. –After a long time, the capacitor behaves like a wire.

Discharging Capacitor C a b + --  R + I I The capacitor is initially fully charged, Q = Q 0. At t = 0 the switch is thrown from position a to position b in the circuit shown. From KVL: Q (t) must be an exponential function of the form: Therefore, where From initial condition, Q(0) = Q 0, we get:

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

–determine which KCL equations are algebraically independent (not all are in this circuit!) –I 1 = I 2 + I 3 –I 4 = I 2 + I 3 –I 4 = I 5 + I 6 –I 1 = I 5 + I 6 Appendix: A three-loop KVL example Identify all circuit nodes - these are where KCL eqn’s are found Analyze circuit and identify all independent loops where  V i = 0  KVL I1=I4I1=I2+I3I4+I5+I6I1=I4I1=I2+I3I4+I5+I6 I1I1 I2I2 I3I3 I4I4 I5I5 I6I6

A three-loop KVL example Now, for Kirchoff’s voltage law: (first, name the resistors) I1=I4I1=I2+I3I4+I5+I6I1=I4I1=I2+I3I4+I5+I6 I1I1 I2I2 I3I3 I4I4 I5I5 I6I6 Here are the node equations from applying Kirchoff’s current law: There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws R1R1 R 2a R 2b R3R3 R4R4 R6R6 R5R5  I 6 R 6 +I 5 R 5 =0 I 2 R 2b +I 2 R 2a - I 3 R 3 =0 V B -I 1 R 1 -I 2 R 2a - I 2 R 2b -I 4 R 4 -I 5 R 5 = 0 Six equations, six unknowns….

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