Chapter 27 Circuits

In this chapter we will cover the following topics:

-Electromotive force (emf) -Ideal and real emf devices -Kirchhoff’s loop rule -Kirchhoff’s junction rule -Multiloop circuits -Resistors in series -Resistors in parallel -RC circuits, charging and discharging of a capacitor

(27 – 1)

A cylindrical copper rod has resistance R. It is reformed to three times its original length with no change of volume. Its new resistance is:

A. RB. 3RC. 9RD. R/9E. R/3

In order to create a current through a resistor, a potential

difference must be created across its terminals. One way

of doing this is to connect the resistor to a battery. A device

that can maintain a potential difference between two terminals

is called a or an . Here emf

stands for: electromotive force. Examples of emf devices:

a battery, an electric generator, a so

"seat of an emf " "emf device"

lar cell, a fuel cell, etc

pump

High (+) reservoir

Low (-) reservoir

These devices act like "charge pumps" in the sense that they move positive charges

from the low potential (negative) terminal to the high potential (positive) terminal.

A mechanical analog is given in the figure below.

In this mechanical analog a water pump transfers

water from the low to the high reservoir. The water

returns from the high to the low reservoir through a

pipe which is the analog of the resistor.

The emf (symbol ) is defined as the potential

difference between the terminals of the emf device

when no current flows through it.

E

(27 – 2)

The polarity of an emf device is

indicated by an arrow with a small circle at its tail.

The arrow points from the negative to the positive

terminal of the device. When the emf device is

co

Notation :

nnected to a circuit its internal mechanism transports

positive charges from the negative to the positive terminal

and sets up a charge flow (a.k.a. current) around the

circuit. In doing so the emf device does work on

a charge which is given by the equation:

. The required energy comes

from chemical reactions in the case of a battery; in the case

of a generator it comes from the mecha

dW q

dq

d

dW

E

nical force that

rotates the generator shaft; in the case of a solar cell it

comes from the sun. In the circuit of the figure

the energy stored in emf device B changes form:

It does mechanical work on the motor. It produces

thermal energy on the resistor. It gets converted into

chemical energy in emf device A

(27 – 3)

An emf device is said to be if the voltage across its

terminals and does depend on the current that flows

through the emf device.

An emf device is s

id

a

V

a b i

V

Ideal and real emf devices

ideal

not

E

to be if the voltage across its

terminals and with current according to

the equation: .

The parameter is known as the device's " "

of the emf device

V ir

V

a b i

r

real

decreases

internal resistance

E

.

V

i

E

Ideal emf device

V

i

E

Real emf device

V E

V ir E

(27 – 4)

Consider the circuit shown in the figure. We assume

that the emf device is ideal and that the connecting

wires have negligible resistance. A current flows

through the c

i

Current in a single loop circuit

ircuit in the clockwise direction.

In a time interval a charge passes through the circuit. The battery is

doing work . Using energy conservation we can set this amount

of work equal to the rate at which heat is

dt dq idt

dW dq idt

E Ei i

2 generated on R.

Kirchhoff put the equation above in the form of a rule known as Kirchhoff's l

0

oop rule

(KLR for short)

idt Ri dt

Ri iR

Ei

E Ei i

0iR Ei

The algerbraic sum of the changes in potential encountered in a complete

traversal of any loop in a circuit is equal to zero.

KLR :

The rules that give us algebraic sign of the charges in potential through a resistor

and a battery are given on the next page.(27 – 5)

Ri

motion

-V iR

Ri

motion

V iR

motion-V E+ -

motionV E+-

For a move through a resistance

in the direction of the current, the change in the

potential

For a move through a resistance in the direction

opposite to that of the current, th

V iR

Resistance Rule :

e change in the

potential V iR

For a move through an ideal emf device

in the direction of the emf arrow,

the change in the potential

For a move through an ideal emf device in a direction

opposi

te to that of the

V

EMF Rule :

E

emf arrow,

the change in the potenti al V E

(27 – 6)

Consider the circuit of fig.a. The battery is real with internal

resistance . We apply KLR for this loop starting at point a and going counterclockwise:

0 . We note tha

r

ir iR iR r

KLR example :

EE t for an ideal battery 0 and

The internal resistance of the battery is an integral part of the battery

internal mechanism. There is no way to open the battery and remove .

In fig.b we

r iR

r

r

Note :

E

plot the potential V of every point in the loop as we start at point a

and go around in the counterclockwise direction. The change in the battery

is positive because we go from the negative to th

Ve positive terminal. The change

across the two resistors is negative because we chose to traverse the loop

in the direction of the current. The current flows from high to low potential

V

(27 – 7)

Consider the circuit shown in the figure. We wish to

calculate the potential difference between

point b and point a.b aV V

Potential difference between two points :

We choose a path in the loop that takes us from the initial point a to the final point b.

sum of all potential changes along the path.

There are two possible paths: We will try them both.

Le

f iV V V

ft path:

Right path:

The values of we get from the two paths are the same.

b a

b a

b a

V V ir

V V iR

V V

Note :

E

sum of all potential changes along the path from point a to point b b aV V V

(27 – 8)

i

V

Consider the combination of resistors

shown in the figure . We can substitute

these combinations of resistors with a

single resistor that is

"electrically equivalent"

to the r

eqR

Equivalent Resistance :

esistor group it substitutes.

This means that if we apply the same voltage across the

resistors in fig.a and across the same current is provided by the battery.

Alternatively, if we pass the same current through the

eq

V

R i

i circuit in fig.a

and through the equivalent resistance , the voltage across them is identical.

This can be stated in the following manner: If we place the resistor

combination and the equivalent

eqR V

resistor in separate black boxes,

by doing electrical mesurements we cannot distinguish between the two.

(27 – 9)

1 2 3

Consider the three resistors connected in series

(one after the other) as shown in fig.a. These resistors have

the but different voltages , , and

The net

i V V Vs

Resistors in ser

ame cu

ies :

rr

ent

e e

1 2 3

1 2 31 2 3

voltage across the combination is the sum

We will apply KLR for the loop in fig.a starting at point a,

and going around the loop in the counterclockwise direction:

0

V V V

iR iR iR iR R R

EE

1

( )

We will apply KLR for the loop in fig.b starting at point a,

and going around the loop in the counterclockwise direction:

0 ( )

If we compare eqs.1 with eqs.2 we get: q

eeq

e

qiR iR

R R

eqs.1

eqs.2E

E

2 3

1 21

For resistors connected in series the quivalent resistance i

..

s:

.

n

eq i ni

R R

R R R

n

R R

1 2 3eqR R R R

(27 – 10)1 2 ...eq nR R R R

Consider the circuit shown in the figure. There

are three brances in it: Branch bad, bcd, and bd.

We assign currents for each branch and define the

current directions arbitrarily.

Multiloop circuits :

The method is self

correcting. If we have made a mistake in the direction

of a particular current the calculation will yield

a negative value and thus provide us with a warning.

1 2 3

1 3 2

1

We asign current for branch bad, current for branch bcd, and current

for branch bd. Consider junction d. Currents and arrive, while leaves.

Charge is conserved thus we have :

i i i

i i i

i 3 2. This equation can be fomulated

as a more general principle knwon as Kirchhoff's junction rule (KJR)

i i

The sum of the currents entering any junction is equal to the sum of the currents

leaving the junction

KJR :

(27 – 11)

1 2 3

1 3 2

In order to determine the currents , , and

in the curcuit we need three equations. The first

equation will cone from KJR at point d:

KJR/junction d: ( )

i i i

i i i eqs.1

1 1 1 3 3

The other two will come from KLR: If we traverse the left loop (bad) starting at b

and going in the counterclockwise direction we get:

KLR/loop bad: 0 ( ) Now we go around the rig hi R i R eqs.2E

3 3 2 2 2

1 2 3

t loop (bcd)

starting at point and going in the counterclockwise direction:

KLR/loop bcd: 0 ( )

We have a system of three equations (eqs.1,2 and 3) and three unknowns , , and

i R i R

i i i

eqs.3E

If a numerical value for a particular current is negative this means that the chosen

direction for this current is wrong and that the current flows in the opposite direction.

We can write a fourth equa

1 1 1 2 2 2

tion (KLRfor the outer loop abcd) but this equation does

not provide any new information.

KLR/loop abcd: 0 ( )i R i R eqs.4E E(27 – 12)

Consider the three resistrors shown in the figure

"In parallel" means that the terminals of the resistors

are connected together on both sides. Thus resistrors

in parallel have the

Resistors in parallel

applied across them.

In our circuit this potential is equal to the emf of the

battery. The three resistors have different currents

flowing through them. The total current is the sum

o

same potential

E

1 2 3 1 2 31 2 3

1 2 3

f the individual currents. We apply KJRat point a.

1 1 1 (eqs.1) From fig.b we have:

(eqs.2) If we compare equations eq

i i i i i i iR R R

iR R R

iR

E E E

E

E

1 2 3

1 1 1 1

1 and 2

we get: eqR R R R

1 2 3

1 1 1 1

eqR R R R

(27 – 13)

An ammeter is an istrument that measures current.

In order to measure the current that flows through

a conductor at a certain point we must cut the

conductor at this point and co

Ammeters and Voltmeters

nnect the two ends

of the conductor to the ammeter terminals so that

the current can pass through the ammeter.

An example is shown in the figure, where

the ammeter has been inserted between points a and b.

1 2

It is essential that the ammeter resistance be much smaller than the other

resistors in the circuit. In our example: and

A voltmeter is an instrument that measures the potential diff

A

A A

R

R R R R

1

erence between two point

in a circuit. In the example of the figure we use a voltmeter to measure the potential

across . The voltmeter terminals are connected to the two points c and d.

It is essenti

R

1 2

al that the voltmeter resistance be much lerger than the other

resistors in the circuit. In our example: and V

V V

R

R R R R (27 – 14)

+

-

iConsider the circuit shown in the figure. We assume

that the capacitor is initially uncharged and that at

0 we throw the switch S from the middle position

to posi

t

RC circuits : Charging of a capacitor

tion a. The battery will charge the capacitor

through the resistor .

C

R

Our objective is to examine the charging process as function of time.

We will write KLR starting at point b and going in the the counterclockwise direction.

0 The current q dq dq q

iR i RC dt dt C

E E 0 If we rearrange the terms

we have: This is an inhomogeneous, first order, linear differential

equation with initial condition: (0) 0. This condition expresses the fact that

a

t

dq qR

dt Cq

t

E

0 the capacitor is uncharged.

(27 – 15)

/

Differential equation:

Intitial condition: (0) 0

Solution: 1 Here:

The constant is known as the "time constant" of the circuit.

If we plot versus we see t

t

dq qR

dt Cq

q C e RC

q t

E

E

hat does not reach its terminal

value but instead increases from its initial value and it

reaches the terminal value at . Do we have to wait for

an internity to charge the capacitor? In pract

q

C

t E

/

ice no.

( ) 0.632

( 3 ) 0.950

( 5 ) 0.993

If we wait only a few time constants the charge, for all

practical purposes has reached its terminal value .

The current If t

q t C

q t C

q t C

C

dqi edt R

E

E

E

E.

Ewe plot i versus t

we get a decaying exponential (see fig.b) (27 – 16)

RC

+

-

i

q

t

qo

O

Consider the circuit shown in the figure. We assume

that the capacitor at 0 has charge and that at

0 we throw the switch S from the middle position

to pos

ot q

t

RC circuits : Discharging of a capacitor

ition b. The capacitor is disconnected from the

battery and looses its charge through resistor .

We will write KLR starting at point b and going in

the counterclockwise direction: 0

Taking int

R

qiR

C

o account that we get: 0 dq dq q

i Rdt dt C

- /

This is an homogeneous, first order, linear differential equation with initial condition:

(0) The solution is: Where . If we plot q versus t we

get a decaying exponetial.

to oq q q q e RC

The charge becomes zero at . In practical terms we

only have to wait a few time constants.

( ) 0.368 , (3 ) 0.049 , (5 ) 0.007

o o o

t

q q q q q q

(27 – 17)

RC

In the diagram R1 > R2 > R3. Rank the three resistors according to the current in them, least to greatest.

A. 1, 2, 3B. 3, 2, 1C. 1, 3, 2D. 3, 1, 3E. All are the same

A. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOT

I. In the diagram, the current in the 3-resistor is 4A. The potential difference between points 1 and 2 is:

A. 0.75VB. 0.8VC. 1.25VD. 12VE. 20V