1.5 V+–

wire

open switch

closed switch

2-way switch

ideal battery

capacitor

resistor

47 F

4.7 k

These circuit elements and many others can be combined to produce a limitless variety of useful devices• Two devices are in series if they

are connected at one end, and nothing else is connected there

• Two devices are in parallel if they are connected at both ends

Ch. 28

Resistors in Parallel and in Series• When resistors are in series, the same

current must go through both of them• The total voltage difference is

• The two resistors act like one with resistance

1 1V IR 2 2V IR

R1 R2

1 2V V V 1 2I R R 1 2R R R

• When resistors are in parallel, the same potential is across both of them

• The total current through them is

• The two resistors act like one with resistance

R1 R2

1 1 2 2V I R I R 1 2I I I 1 2

V V

R R

VR

I

1

1 2

1 1

R R

1 2

1 1 1

R R R

Warmup 10x

q1

Parallel and Series - Formulas

Capacitor Resistor Inductor*

Series

Parallel

Fundamental Formula

1 2R R R

1 2

1 1 1

R R R 1 2C C C

1 2

1 1 1

C C C 1 2L L L

1 2

1 1 1

L L L

QV

C V IR

L

dIL

dtE

* To be defined in a later chapter

The Voltage Divider

+–

• Many circuits can be thought of as a voltage divider• Intentionally or unintentionally

E

R1

R2

If Mr. Curious has a resistance of 10 k and the light bulb has a resistance of 240

, how bright is Mr. Curious?

1 2R R R

1 2

IR R

E

11 1

1 2

RV IR

R R

E

22 2

1 2

RV IR

R R

E

What’s the voltage drop across each of the resistors?

+–120 V curious

10000117 V

10240V E=

The larger resistor gets most of the voltage

Not very bright

CT – 1 Consider two identical resistors wired in series (one behind the other). If there is an electric current through the combination, the current in the second resistor is A. equal to B. half C. smaller than, but not necessarily half the current through the first resistor. CT – 2 As more identical resistors R are added to the parallel circuit shown here, the total re-sistance between points P and QB

A. increases. B. remains the same. C. decreases.

Ans A

Ans C

CT – 3 Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected,

A. Practically all the charge continues to flow through the bulb. B. half the charge flows through the wire, the other half continues through the bulb. C. Practically all the charge flows through the wire. D. none of the above

Ans C

CT - 4-The circuit below consists of two identical light bulbs burning with equal brightness and a single 12 V battery. When the switch is closed, the brightness of bulb A

A. increases. B. remains unchanged. C. decreases. CT - 5-If the four light bulbs in the figure are identical, which circuit puts out more light?

A. I. B. The two emit the same amount of light. C. II.

Ans A

Ans A

Ex- (a) Find the equivalent resistance between points a and b in the figure below. (b) If a potential difference of 34 V is applied between points a and b, calculate the current in

each resistor. Solve on Board

Ideal vs. Non-Ideal Batteries• Up until now, we’ve treated a

battery as if it produced a fixed voltage, no matter what we demand of it

• Real batteries also have resistance• It limits the current and therefore

the power that can be delivered• If the internal resistance r is small

compared to other resistances in the problem, we can ignore it

ideal battery

+–

realistic battery

E

+–Er

The maximum potential difference E across the

battery is called electromotive force (emf)+–30 V10

50 A 30 V battery with 10 of resistance is connected to a 50 resistor. What is the actual voltage across the 50 resistor?A) 30 V B) 36 V C) 6 VD) 25 V E) 24 V

1r

Vr R

E 25 V

JIT

Ans: (iii) a (ii) bAns: (i) b (ii) a

Kirchoff’s First LawThe total current into any vertex equals the current out of that vertex

How to apply it:• First, assign a current and a direction to every

pathway• Two components in series will always have the

same current• At every vertex, write the equation:

in outI I

+–

+ –

3

5

4

12 V

6 V

I1

I3

ABWhich equation do you get for point A?

A) I1 + I2 = I3 B) I2 + I3 = I1 C) I1 + I3 = I2 D) I1 + I2 + I3 = 0

• The equation from point B is

3 1 2I I I You always get oneredundant equation

I2

Kirchoff’s Second Law The total voltage change around a loop is always zero

How to apply it:• First, assign a direction to every loop

• I often pick clockwise• Start anywhere, and set 0 equal to sum of

potential change from each piece:• For batteries: V = E

• It is an increase if you go from – to +• It is a decrease if you go from + to –

• For resistors: V = IR• It is a decrease if you go with the current• It is an increase if you go against the current

+–

+ –

3

5

4

12 V

6 V

I1

I2

I30 12 13I 6 25I

1 20 18 3 5I I

Kirchoff’s Second Law (2)

+–

+ –

3

5

4

12 V

6 V

I1

I2

I3

What is Kirchoff’s Second Law for the purple loop?A) 0 = +5I2 – 6 – 4I3 B) 0 = +5I2 + 6 – 4I3

C) 0 = –5I2 – 6 – 4I3 D) 0 = –5I2 + 6 – 4I3

20 5I 6 34I 3 1 2I I I

1 20 18 3 5I I

Three equations in three unknowns: solve it• We can let Maple do it for us

> solve({i3=i1+i2,0=-5*i2-6.-4*i3,0=18-3*i1+5*i2},[i1,i2,i3]);

Negative currents means we guessed the wrong way• Not a problem

Ex- Using Kirchoffs' rules, (a) find the current in each resistor shown below. (b) Find the potential difference between points c and f. Which point is at a higher potential?

Solve on Board

Kirchoff’s Laws with Capacitors

• The voltage change is given by V = Q/C• It is a decrease if (+)Q is the side you are going in• It is an increase if Q is the side you are going out

• The current is related to the time change of Q• Add minus sign if I doesn’t enter from the same side

as Q – it is minus if decreasing• If you are in a steady state, the current through a

capacitor is always zero

I

C

Q

+–

dQI

dt

+–+

–

In this circuit, in the steady state, where is

current flowing?

It’s really just a battery and two resistors in series!

If know which side is positive then that is high potential – work like battery.

The Simplest RC Circuit

Q0

C

RIn the circuit shown at left, the capacitor starts

with charge Q0. At time t = 0, the switch is closed. What happens to the charge Q?

I

• Current begins to flow around the loop, so the charge Q will change

0Q

RIC

dQ

Idt

Q

RC

• This is a differential equation, and therefore hard to solve

dQ dt

Q RC

dQ dt

Q RC ln

tQ k

RC

t RCQ e

0t RCQ Q e

Check the units:

FRC V C

A V

C

C s s

Charging and Discharging Capacitors

0tQ Q e

0t RCQ Q e

RC

• The combination RC = is called the time constant• It’s the characteristic time it takes to discharge

• We can work out the current from

dQI

dt 0 tQ e

In this circuit, the capacitor is initially uncharged, but at t = 0 the switch is closed

Q

C R +– E

I

dQI

dt

0Q

IRC

E

dQ Q

dt RC R

E

1 t RCQ C e E

0 0t tC e eRC R

Warmup 11

CT – 6 A simple circuit consists of a resistor R, a capacitor C charged to a potential Vo (not shown), and a switch that is initially open but then thrown closed. Immediately after the switch is thrown closed, the current in the circuit is

A. V o /R. B. zero.

C. need more information

Ans A

JIT Quick Quiz 28.5Consider the circuit in the figure and assume that the battery has no internal resistance.

(i) Just after the switch is closed what is the current in the battery (a) 0

(b) e/2R,(c) 2e/R, (d) e/R, (e) impossible to determine

(ii) After a long time, what is the current in the battery?

Ans i) c, ii) d

Example (Serway 28-38). Consider a RC circuit consisting of a Emf = 30.0V, a resistor = 1.00 Ma capacitor = 5.00 F and a switch like in Figure 28.34 (charging a capacitor). Find (a) the time constant of the circuit and (b) the maximum charge on the capacito after the switch is closed. (c) If the switch is closed at t = 0, find the current in the resistor 10.0 s later.

Solve on Board

Ammeters and Voltmeters• An ammeter is a device that measures the current (amps) anywhere

in a circuit• To use it, you must route the current through it• A perfect ammeter should have zero resistance

• A voltmeter is a device that measures the potential difference (volts) between any two points in a circuit• To use it, you can simply connect to any two points • A perfect voltmeter has infinite resistance

V

A

+–

A

V

V A

Which meter is installed incorrectly?A) Left voltmeter B) Right voltmeterC) Left ammeter D) Right ammeterE) All are correct

• Voltmeters should be connected to two places in an existing circuit• The left voltmeter is placed correctly

• A voltmeter has infinite resistance• The right one effectively blocks the current on the right

Household Wiring• All household appliances consume electrical power

• Think of them as resistors with fixed resistance R• Devices are designed to operate at 120 V*

• Often, they give the wattage at this voltage• Can easily get the effective resistance from

• To make sure power is given to each device, they are all placed in parallel

+–

A

2V R P

*Actually, this is alternating current, later

chapter

• If you put too many things on at once, a lot of current is drawn• The wires, which have some resistance, will start to get hot• To avoid setting the house on fire, add a fuse (or a circuit

breaker)

Fuse

box

Inside House

Warmup 11

Why three wires?• If a device is functioning properly, you need only two wires

• “Live” and “Neutral” wires

Toaster

• If the live wire accidentally touches the casing, the person can be electrocuted

• The wrong solution – connect the neutral to the casing• Now imagine the neutral wire breaks

• The person again can be electrocuted• The right solution: Add a third “ground” wire connected

directly to ground• Normally no current will flow in this wire• If the hot wire touches the casing, it will trigger the

fuse/circuit breaker and protect the person