26 lecture lam - university of hawaiiplam/ph272_summer/l6/26_lecture_lam.pdf · introduction • in...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 26

Direct-Current Circuits

Modified by P. Lam 7_23_2008


Topics for Chapter 26

• Resistors in series and parallel circuit.

• Kirchoff’s Rules.

Intermission

• R-C circuits


Introduction

• In the last chapter, we gainedinsight about how current flowsthrough a resistor in simpleexamples like a light bulbattached to a battery.

• Now, imagine many thousandsof circuits wired onto flatwafers with structure so tinythat microscopy would benecessary to view their patterns.Understanding the next step andmastering more complex circuitpatterns is the goal forChapter 26.


Resistors in series and parallel I

• If we took three resistors

and considered the

different ways they

could be connected, we

arrive at the four

possibilities illustrated

in Figure 26.1.


Equivalent resistance for series and parallel circuits

Series resistors :

(1) Same current through all resistors = I

(2) Total voltage V = V1 +V2 +V2

V = IR1 + IR2 + IR3 = IReq

Req = R1 + R2 + R3

Parallel resistors :

(1) Same voltage across all resistors = V

(2) Total current I = I1 + I2 + I2

I =V

R1

+V

R2

+V

R3

=V

Req

1

Req

=1

R1

+1

R2

+1

R3

Note : Formula for combining resistance

are "opposite" to those for combining capacitance.

Reason : V = IR but V = Q1C


Examples of Resistors in series and parallel

• If you have ever wired a Christmas tree with a series of lights

(resistors) in series, you know what happens if just one burns out.

The lights have become an open circuit and will not function.

• Car headlights are a good example of resistors wired in parallel. If

one light burns out, the circuit changes but still functions to allow

the driver a safe trip to repair. See Figure 26.2 below.


Example of Resistors in series and parallel—combinations

Follow Example 26.1, find I through each resistor.


Equivalent resistance method is not always useful

• The two circuits below CANNOT be reduced to simple series-parallel combination of resistors.

(a) When there are more than

one battery

(b) The arrangement of resistors

is simply neither series nor

parallel (the spoiler is middle

resistor).


Kirchoff’s Rules I—junctions

• Kirchoff’s Rules: (1) Junction rule and (2) Loop rule can be usedto find current and voltages for any circuit.

Note:(1) Junction rule = conservation of charge

(2) You may ASSUME the direction of the current; if your final anwer is then the

actual current flows in the opposite direction.


Kirchoff’s Rules II—loops

• The algebraic sum of thepotential differences inany loop must equal zero.

• Example:

I1 I2I3

Loop 1 in fig. a (take c as the starting point)

+ 1 I1r1 + I2R = 0

Loop 2 : (take c as the starting point)

+ 1 I1r1 I3r2 2 = 0

Loop 3 : (take b as the starting point)

You try it!

Suppose all the emf and resistance are given.

The three unknowns are I1,I2,and I3.

We have 3 loop equations plus 1 junction equation

I1 + I2 = I3

Do we have too many equations?


Kirchoff’s Rules - convention for potentials in loop equation.

Think of direction of current as

direction of water flow in a

water fall, which flows from

high gravitational potential to

low gravitational potential


Kirchoff’s Rules—examples

Find the unknowns: , I, and r


Kirchoff’s Rules—examples

Find I1, I2, and I3


Power distribution systems—a home

• Potential, resistors, outlets, input from the power company … no

wonder electricians are integral contractors in home construction!


Intermission


R-C circuits -charging (qualitative)

Sketch the qualitative behavior of Vc(t), VR(t), I(t).

(Let the switch be closed at t-0)

Label your graphs with numerical values;

given: =10V,R=5 ,C=4F, initial charge on capacitor=0.


RC-circuit - discharging (qualitative)

Follow Examples 26.12 and 26.13.

Express Io in terms of Qo, R and C


RC-circuit - discharging (quantitative)

• Follow Examples 26.12 and 26.13.

Kirchoff’s loop rule: a ->b->c->a

IRQ

C= 0 Note : I =

dQ

dt

(Based on the direction of the current, I = positive

but dQ/dt is negative)

RdQ

dt+Q

C= 0

dQ

dt+Q

RC= 0

dQ

dt=

Q

RC

This is called "first - order, linear, homegeneous differentia equation" (D.E.).

Think of a function whose derivative is

a constant times the function?

Q(t) = AeBt

Substitute into the D.E. B = -1

RCA is set by the initial condition, A = Q(t = 0) = Qo

Q(t) = Qoet /RC ; RC = time constant

I(t) =dQ

dt=

Qo

RCe t/RC

(Compare Q(t) and I(t) with the graphs on the previous slide)

Vc (t) =Q(t)C

=Qo

Ce t /RC and VR = IR =

Qo

Ce t /RC


RC-circuit - charging (quantitative)

Kirchoff’s loop rule: a ->b->c->aIR

Q

C+ = 0 IR +

Q

C=

Note :Now I = +dQ

dt (I is positive and

dQ

dt is positive)

RdQ

dt+Q

C=

This is called "first - order, linear, inhomegeneous differentia equation" (D.E.).

The solution to this D.E. has two parts :

inhomogeneous solution (or called steady state solution) +

homogenous solution (or called transient solution)

Steady state solution : (i.e. dQ/dt = 0) Qsteady = C

Homogeneous solution : Q(t)homogeneous = Ae t /RC

Complete solution : Q(t) =C + Ae t /RC

Again, A is set by the initial condition.

Q(t = 0) = Qo = C + A A =Qo C

For the case where Qo = 0, then A = -C

Q(t) =C [1 e t /RC ]

I(t) =dQdt

=CRC

e t/RC =R

e t/RC

(Compare Q(t) and I(t) with the graphs on the earlier slide)

Vc (t) =Q(t)

C and VR = I(t)R


Complicated RC-circuit

Given : Initially the capacitor is uncharged.

It is not easy to analyze this circuit to find currents and voltages as a function of time.

However, it is very straight forward to find the current and voltages at t = 0 (right after the switch is closed)

and when the circuit reaches steady state.

(a) t = 0+. Initially the capacitor is uncharged VC = 0

The capacitor acts like a "short circuit"

can replace the capacitor by a piece of conducting wire.

Circuit contains only resistors which we can find Req and currents.

(b) t (steady state). The capacitor is charged and there is no

more current going through the capacitor

the capacitor acts like an "open circuit"

I = 0 for the circuit branch which contains the capacitor

we can remove that branch from the circuit.

Again, find Req and currents.

26 lecture lam - university of hawaiiplam/ph272_summer/l6/26_lecture_lam.pdf · introduction • in...

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