18. Electric Current & Direct Current Circuits

SF026 2015/ 16

ANALOGY: Electric circuits

Voltage: A force

that pushes the

current through the

circuit (in this picture

it would be

equivalent to gravity)

Resistance:

Friction that

impedes flow of

current through

the circuit (rocks

in the river)

Current:

The actual

“substance”

that is

flowing

through the

wires of the

circuit

(electrons!)

18.1 Electrical Conduction

(a)Describe microscopic model of current

(b)Define and use electric current formulae,

dt

dQI

Microscopic Model of Current

In wire (metal) the charge carrier is free electrons

Without battery, V = 0 No

electric field, E

Free e undergoes

random motion

Microscopic Model of Current

When a potential difference is

applied across the metal, an

electric field, E is set up

This field exerts an electric force

on the freely moving electron

The freely moving electrons tend to drift with constant

average velocity (drift velocity, vd) along the metal in a direction opposite that of

the E

http://www.schoolphysics.co.uk/age16-

19/Electricity%20and%20magnetism/Current%20electricity/text/Electric_current/index.html

Electric Current, I

Consider a simple closed circuit consists of wires, a battery and a light bulb as shown

From the diagram,

Direction of electric field or electric current: (+)ve to (–)ve

Direction of electron flows: (–)ve to (+)ve

Electric current, I is defined as the rate of flow of chargeMathematically,

One Ampere is the flow of one C through an area in one second

t

QI

dt

dQI Instantaneous

current

Average

current

It is a base and scalar quantityS.I. unit: ampere (A).

1s C 1second 1

coulomb 1ampere 1

Example 18.1

(a) There is a current of 0.5

A in a flashlight bulb for

2 min. How much charge

passes through the bulb

during this time?

Solution:

(b) A silver wire carries a

current of 3.0 A.

Determine

(i) the number of electrons

per second pass

through the wire,

(ii) the amount of charge

flows through a cross-

sectional area of the

wire in 55 s.t

QI C60

119 s electrons 1088.1 t

N

C 165Q

18.2 Ohm’s Law & Resistivity

(a)State and use Ohm’s law

(b)Define and use resistivity,l

RAρ

Ohm’s Law

States that the potential

difference across a

conductor is

proportional to the

current flowing through

it if its temperature is

constant

where T is constant

Materials that obey Ohm’s law

are materials that have constant

resistance over a wide range of

voltage ohmic conductor

Materials that do not obey the

Ohm’s law non-ohmic

conductors

IV

IRV Then

R = resistance

(V)V

(A)I0

Gradient, m

= R

Ohmic conductor

(metal)

http://www.schoolphysics.co.uk/animations/Ohms_law/index.html

file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-ohms-law/ohms-law_en.html

Ohm’s Law – Non-ohmic conductor

V

I0

Semiconductor Carbon

V

I0

Electrolyte

V

I0

Resistance, R – electric property which is

impedes or limits current in an electrical circuit

is defined as a ratio of the potential difference across an electrical component to the current passing through it

Mathematically,

Scalar quantity, unit: ohm ( ) or V A1

I

VR

Resistance

Type of material

Length, lCross

sectional area, A

Temperature

In general, the

resistance of

a conductor

increases

with

temperature

V: potential difference (voltage),I: current

Resistivity &

Conductivity

Resistivity, (specific resistance)

is defined as the resistance of a unit cross-sectional area per unit length of the materialMathematically,

Scalar quantity Unit: ohm meter ( m) It is a measure of a material’s

ability to oppose the flow of an electric current

Resistivity depends on the type of the material & temperature

l

RAρ

ρσ

1

Conductivity,

is defined as the reciprocal of the resistivity of a material

Mathematically,

Scalar quantity,

unit: 1 m1

Material Resistivity, ( m)

Silver 1.59 108

Copper 1.68 108

Aluminum 2.82 108

Gold 2.44 108

Glass 10101014

A good electric conductors

have a very low resistivities,

good insulators have very

high resistivities

file:///C:/DOCUME~1/KMS/LOCALS~1/Temp/phet-resistance-in-a-wire/resistance-in-a-wire_en.html

Example 18.2

A constantan wire of

length 1.0m and cross

sectional area of 0.5 mm2

has a resistivity of 4.9 x 10–7 Ωm. Find the

resistance of the wire.

Example 18.3

Two wires P and Q with circular cross section are made of the same metal and have equal length. If the resistance of wire P is three times greater than that of wire Q, determine the ratio of their diameters.

Exercise 18.2

1. A wire 5.0 m long and 3.0 mm in diameter has a

resistance of 100 . A 15 V of potential difference is

applied across the wire. Determine

(a) the current in the wire,

(b) the resistivity of the wire,

(c) the rate at which heat is being produced in the wire.

(College Physics,6th edition, Wilson, Buffa & Lou, Q75,

p.589)ANS: 0.15 A; 1.414 104 m; 2.25 W

18.3 Variation of Resistance

With Temperature

(a)Explain the effect of temperature on

electrical resistance

(b)Use resistance,

00 1 TTRR

Effect of Temperature on Electrical

Resistance in Metals

When the temperature increases, the number of free electrons per unit volume in metal remains unchanged

Metal atoms in the crystal lattice vibrate with greater amplitude and cause the number of collisionsbetween the free electrons and metal atoms increase and slowing down the electron flow

Hence the resistance in the metal increases

Effect of Temperature on Electrical

Resistance in Metal

The resistance of a metal

can be represented by the

equation below

whereR = final resistance

Ro= initial resistance

= the temperature coefficient of resistivity

Material (C1)

Silver 4.10 103

Mercury 0.89 103

Iron 6.51 103

Aluminum 4.29 103

Copper 6.80 103

TαRR 10

α is defined as the fractional change in resistance per Celsius degree

temperature coefficients of resistivity for various materials

T

RR

0

Effect of Temperature on Electrical

Resistance in Metal

19

R

T0

0R

cT

Figure 18.8a : metal Figure 18.8b : semiconductor

R

T0

R

T0Figure 18.8c : superconductor

R

T0Figure 18.8d : carbon

Example 18.4

A copper wire has a resistance of 25 m at 20 C. When

the wire is carrying a current, heat produced by the current

causes the temperature of the wire to increase by 27 C.

(a) Calculate the change in the wire’s resistance

(b) If its original current was 10.0 mA and the potential

difference across wire remains constant, what is its final

current?

(copper = 6.80 103 C1)

Example 18.5

A platinum wire has a resistance of 0.5 Ω at 0°C.

It is placed in a water bath

where its resistance rises to a final value of 0.6 Ω.

What is the temperature

of the bath?

(platinum = 3.93 103

C1)

Exercise 18.3

1. A wire of unknown composition has a resistance of 35.0

when immersed in the water at 20.0 C. When the wire is

placed in the boiling water, its resistance rises to 47.6 .

Calculate the temperature on a hot day when the wire has a

resistance of 37.8 . (Physics,7th edition, Cutnell &

Johnson, Q15, p.639)

ANS: 37.78 C

2. A copper wire has a resistance of 25 m at 20 C. When the

wire is carrying a current, heat produced by the current

causes the temperature of the wire to increase by 27 C.

(a) Calculate the change in the wire’s resistance.

(b) If its original current was 10.0 mA and the potential

difference across wire remains constant, what is its final

current? (copper = 6.80 103 C1)ANS: (a) 4.59×10–3 Ω; (b) 8.45×10–3 A

18.4 Electromotive Force (emf),

Potential Difference & Resistance

(a)Define emf, of a battery

(b)Explain the relationship between emf of

a battery and potential difference across

the battery terminals

(c)Use terminal voltage,

IrεV

E.m.f, VS Potential Difference, V

Terminal voltage, V is the potential difference across the terminals of a battery when there is a current flowing through it

Electromotive force, e.m.f (ξ) of a battery is the maximum potential difference across its terminals when it is not connected to a circuits

Emf in electric

source, p.d

between two

points in a circuit

E.m.f, VS Potential Difference, V

Consider a circuit

consisting of a battery

(cell) that is connected

by wires to an external

resistor R as shown in

Figure

In reality, when a battery

is supplying current, its

terminal voltage is less than its e.m.f, ξ

This reduces of voltage

is due to energy

dissipation in the battery

In effect, the battery has

internal resistance (r)

E.m.f, VS Potential Difference, V

Mathematically

V = IR,

where = emf

V = terminal potential difference

I = current

R = external resistance

r = internal resistance

is the resistance due to chemicals inside the battery (cell)

It will constitutes part of the total resistance in a circuit

The emf of a battery is constant but the internal resistance of the battery increases with time as a result of chemical reaction

I R r

V Ir

Internal resistance, r

rε r εOR

Example 18.6

A battery has an emf of 9.0

V and an internal resistance

of 6.0 . Determine

(a) the potential difference

across its terminals

when it is supplying a

current of 0.50 A,

(b) the maximum current

which the battery could

supply.

Example 18.7

A car battery has an emf of

12.0 V and an internal

resistance of 1.0 . The

external resistor of

resistance 5.0 is

connected in series with the

battery as shown.

Determine the reading of

the ammeter and voltmeter

if both meters are ideal.

Exercise 18.4

1. A battery of e.m.f 3.0 V and internal resistance 5.0 is connected to a switch by a wire of resistance 100 . The voltage across the battery is measured by a voltmeter. What is the voltmeter reading when the switch is

(a) off?(b) on? ANS: (a) 3V, (b) 2.86 V

2. An idealized voltmeter is connected across the terminals of a battery while the current is varied. Figure shows a graph of the voltmeter reading V as a function of the current I through the battery. Find

(a) the emf, ξ and(b) the internal resistance of the battery

ANS: (a) 9V, 4,5Ω

Exercise 18.4

3. A battery of emf 6.0 V is connected across a 10

resistor. If the potential difference across the resistor is

5.0 V,

(a) Determine

(i) the current in the circuit,

(ii) the internal resistance of the battery.

(b) When a 1.5 V dry cell is short-circuited, a current of 3.0

A flows through the cell. What is the internal resistance

of the cell?ANS: 0.50 A, 2.0 ; 0.50

Revision

ItQ

IV

Laws'Ohm

l

RAρ

ρσ

1

TRR 10

I Battery (cell)

A Brε

R

I R r

IRV

Use:

(i) power,

(ii) electrical energy,

IVP

IVtW

18.5 Electrical Energy & Power

Power, P

is defined as the energy liberated per unit time in the electrical device

The electrical power Psupplied to the electrical device is given by

When the electric current flows through wire or passive resistor, hence the potential difference across it is

then the electrical power can be written as

It is a scalar quantity and its unit is watts (W)

t

VIt

t

WP

IVP

IRV

RIP 2 OR

R

VP

2

Energy, E

Consider a circuit

consisting of a battery that

is connected by wires to

an electrical device (such

as a lamp, motor or battery

being charged) where the

potential different across

that electrical device is V

A current I flows from the terminal A to the terminal B, if it flows for time t, the charge Q which it carries from B to A is given by

Then the work done on this charge Qfrom B to A (equal to the electrical energy supplied) is

If the electrical device is passive resistor(device which convert all the electrical energy supplied into heat), the heat dissipated H is given by

Electrical device

A B

VI I

QVW

ItQ

VItEW

VItWH OR RtIH 2

Example 18.7

In figure below, a battery

has an emf of 12 V and an

internal resistance of 1.0 .

Determine

(a) the rate of energy

transferred to electrical

energy in the battery,

(b) the rate of heat

dissipated in the battery,

(c) the amount of heat loss

in the 5.0 resistor if the

current flows through it

for 20 minutes.

Exercise 18.5

An electric toy of resistance 2.50 is operated by a dry cell

of emf 1.50 V and an internal resistance 0.25 .

(a) What is the current does the toy drawn?

(b) If the cell delivers a steady current for 6.00 hours,

calculate the charge pass through the toy.

(c) Determine the energy was delivered to the toy.

ANS: 0.55 A; 1.19 104 C; 16.3 kJ

18.6 Resistors in Series

& Parallel

Derive and determine effective resistance

of resistors in series and parallel

Resistors in

Series

From the definition of resistance, thus

Substituting for V1, V2 , V3

and V

where

Reff: effective) resistance

;22 IRV ;33 IRV ;11 IRV effIRV

321eff IRIRIRIR

321eff RRRR

Consider three resistors are

connected in series to the

battery

Characteristics:

Same current I flows through each resistor

Total potential difference, V (Assumption: the connecting wires have no resistance)

321 IIII

321 VVVV

Example

(a) What is the current in each resistor?

(b) What is the voltage across each resistor?

(c) What is the total resistance?

(d) What is the battery voltage?

ANS: 0.1A, 3V, 4V, 5V; 120; 12V

Resistors in

Parallel

Consider three resistors

are connected in parallel

to the battery

Characteristics:

Same potential difference, V across each resistor

Total current in the circuit

From the definition of resistance, thus

Substituting for I1, I2 , I3 and I321 VVVV

321 IIII

;2

2R

VI ;

3

3R

VI ;

1

1R

VI

effR

VI

321eff R

V

R

V

R

V

R

V

321eff

1111

RRRR

Example

(a) What is the total resistance of the circuit (watch

out for the bear trap)?

(b) What is the current through each resistor?

(c) What is the total current?

ANS: 12.8; 0.40A, 0.30A, 0.24A; 0.94A

Example 18.8

What is the equivalent

resistance of the resistors

in figure below?

R1= R2= R3= R4= 1 Ω

A

BR

1

R

2

R

3

R

4

A

B

R1

R2

R34

A

B

R1

R2

R3R4

5

3ER

A

B

R1R234

Example 18.9

Find the current in & voltage of the 10 Ωresistor shown below.

Series VS Parallel

When connected to the

same source, two light

bulbs in series draw

less power and glow

less brightly than when

they are in parallel

In the series case the same

current flows through both

bulbs.

If one of the bulbs burns out,

there will be no current at all in

the circuit, and neither bulb will

glow

In the parallel case the potential difference

across either bulb remains equal if one of

the bulbs burns out. The current through

the functional bulb remains equal and the

power delivered to that bulb remains the

same . This is another of the merits of a

parallel arrangement of light bulbs: If one

fails, the other bulbs are unaffected. This

principle is used in household wiring

systems

Example 18.10

For the circuit below, calculate

(a) the effective resistance of the circuit,

(b) the current passes through the 12 resistor,

(c) the potential difference across 4.0 resistor,

(d) the power delivered by the battery.

The internal resistance of the battery may be ignored.

Example 18.11

For the circuit above,

calculate the effective

resistance between the

points A and B.

Exercise 18.6

1. Determine the equivalent resistances of the resistors

below.

ANS: 0.80 ; 2.7 ; 8.0

0.2

0.2

0.2

0.2

0.6

01

0.6 0.4

18

16

0.8

0.9

16

0.6

20

Exercise 18.6

2. The circuit below includes a battery with a finite internal resistance, r = 0.50 .

(a) Determine the current flowing through the 7.1 and 3.2 resistors.

(b) How much current flows through the battery?

(c) What is the potential difference between the terminals of the battery? (Physics,3th edition, James S. Walker, Q39, p.728)

ANS: 1.1 A, 0.3 A; 1.4 A; 11.3 V

Exercise 18.6

3. Four identical resistors are connected to a battery as shown below. When the switch is open, the current through the battery is I0.

(a) When the switch is closed, will the current through the battery increase, decrease or stay the same? Explain.

(b) Calculate the current that flows through the battery when the switch is closed, Give your answer in terms of I0. (Physics,3th edition, James S. Walker, Q45, p.728)

ANS: U think

Exercise 18.6

4. Figure below shows the arrangement of five equal resistors in a circuit. Calculate

(a) the equivalent resistance between point xand y.

(b) the voltage across point b and c.

(c) the voltage across point c and y.

18.7 Kirchhoff’s Law

State and use

Kirchhoff’s law

1st Kirchhoff’s Law (KCL)

the sum of the currents entering any junctions in a

circuit must equal the sum of the currents leaving

that junction

outin II

1I

2I3I

321 III 123 III

3I 2I

1I

2nd Kirchhoff’s Law (KVL)

in any loop, the sum of emf is equal to the sum of the

products of current and resistance

For emf,

For product of IR

ε

ε

direction of loop

+-

ε-

ε+

direction of loop

IR

direction of loop

I

RIR

I

R

direction of loop

IR

Example 18.12

For the circuit

below, determine

the current and its

direction in the

circuit.

1.15

.226

50.8 2 ,V 1.51

4 ,V 5.01Loop 1

I

I

II

Example 18.13

For the circuit below, determine

(a) the currents I1, I2 and I,

(b) the potential difference across the 6.7 resistor,

(c) the power dissipated from the 1.2 resistor.

8.9

9.3

V .09V 21

7.6

.21

I1I 2I

Example 18.14

Find the value of current, I , resistance R & the e.m.f, ξ

ξ

D

A

F E

B

IR

1

3

AI 11

AI 22

C

V12

Exercise 18.7

1. For a circuit below, given

1= 8V, R2= 2 , R3= 3

, R1 = 1 and I = 3 A.

Ignore the internal

resistance in each

battery. Calculate

(a) the currents I1 and I2.

(b) the emf, 2.

ANS: 1.0 A, 4.0 A; 17 V

Exercise 18.7

2. Determine the current in each resistor in the circuit below. (College Physics,6th edition, Wilson, Buffa & Lou, Q57, p.619)

ANS: 3.75 A; 1.25 A; 1.25 A

Exercise 18.7

3. Referring to the circuit above, calculate the current I1, I2and I3

ANS: I1=0.66A, I2 = I3 = 0.33A

18.8 Potential Divider

(a) Explain the principle of a potential divider.

(b) Apply equation of potential divider,

VRR

RV

21

11

Potential Divider

A potential divider produces an output voltage that is a fraction of the supply voltage V

This is done by connecting two resistors in series V

1V

1RI

2V

2RI 11 IRV V

RR

RV

21

11

VRR

RV

21

22

Since the current flowingthrough each resistor is the same,

Therefore, the potential difference (voltage) across R1 is given by

Similarly,

21eff RRR

effR

VI an

d

21 RR

VI

Potential Divider

Resistance R1 and R2 can

be replaced by a uniform

homogeneous wire

The potential difference

(voltage) across the wire

with length l1 is

Similarly,

V

I

2l1lI

BA C

2V1V

Vll

lV

21

11

Vll

lV

21

22

Example 18.15

For the circuit below,

(a) calculate the output

voltage

(b) If a voltmeter of

resistance 4000 is

connected across the

output, determine the

reading of the voltmeter.

18.8 Potentiometer &

Wheatstone Bridge

(a) Explain principles of potentiometer and

Wheatstone Bridge and their applications

(b) Use related equations for potentiometer,

and for Wheatstone Bridge,

x

3

2

1

R

R

R

R

2

1

2

1

l

l

ε

ε

Potentiometer

Consider a potentiometer circuit below

The potentiometer

is balanced when

the jockey is at

such a position on

wire AB that there is

no current through

the galvanometer

Thus

Galvanometer

reading = 0ACx VV

Potentiometer

When the potentiometer

in balanced, the unknown

voltage (potential

difference being

measured) is equal to

the voltage across AC

Application: Compare the emfs of two

cells or find unknown emf

o In this case, a potentiometer is set up as illustrated in Figure below, in which AB is a wire of uniform resistance and J is a sliding contact (jockey) onto the wire.

o An accumulator Xmaintains a steady current I through the wire AB

X

BAI

G

I

(2)

(1)

2εS

II

1ε

J

Application: Compare the emfs of two cells

or find unknown emf

Initially, switch S is

connected to the terminal (1)

and the jockey moved until

the emf ξ1 exactly balances

the potential difference (p.d.)

from the accumulator

(galvanometer reading is

zero) at point C. Hence

where

and

then

X

BAI

G

I

(2)

(1)

2εS

II

1ε

C

J

1l

1 ACV

ACAC IRV A

ρlR 1

AC

11

ρlI

A

(1)

Application: Compare the emfs of two cells

or find unknown emf

X

BAI

G

I

(2)

(1)

2εS

II

1ε

C

J

D1l2l

After that, the switch S is

connected to the terminal (2)

and the jockey moved until

the emf ξ2 balances the p.d.

from the accumulator at

point D

Hence

where

and

then

2 ADV

ADAD IRV A

ρlR 2

AD

22

ρlI

A

(2)

Application: Compare the emfs of two cells

or find unknown emf

By dividing eq. (1) and eq. (2) then

Since

Equation above can be written as

1 1

2 2

l

l

1

1

22

ρlI

A

ρlI

A

1ρlR R l

A

1 1

2 2

R

R

Example 18.16

Consider a potentiometer

with a standard battery with

an e.m.f. of 1.0186 V is

used in the circuit. When the resistance is 36 Ω, the

galvanometer reads zero. If

the standard battery is

replaced by an unknown

e.m.f. the galvanometer

reads zero when the

resistance is adjusted to 48 Ω. What is the value of the

unknown e.m.f. ?

Exercise 18.8(a)

1. In figure below, PQ is a uniform

wire of length 1.0 m and

resistance 10.0 . ξ1 is an

accumulator of emf 2.0 V and

negligible internal resistance.

R1 is a 15 resistor and R2 is a

5.0 resistor when S1 and S2

open, galvanometer G is

balanced when QT is 62.5 cm.

When both S1 and S2 are

closed, the balance length is

10.0 cm. Calculate

(a) the emf of cell ξ2

(b) the internal resistance of cell ξ2.

Exercise 18.8(a)

2. Cells A and B and centre-zero

galvanometer G are connected to a

uniform wire OS using jockeys X and Y as

below. The length of the uniform wire OS is

1.00 m and its resistance is 12 . When

OY is 75.0 cm, the galvanometer does not

show any deflection when OX= 50.0 cm. If

Y touches the end S of the wire, OX = 62.5

cm when the galvanometer is balanced.

The emf of the cell B is 1.0 V. Calculate

(a) the potential difference across OY when

OY = 75.0 cm,

(b) the potential difference across OY when Y

touches S and the galvanometer is

balanced,

(c) the internal resistance of the cell A,

(d) the emf of cell A.

Wheatstone Bridge

It is used to measured the

unknown resistance of the

resistor

Figure below shows the Wheatstone bridge circuit consists of a cell of emf , a galvanometer , know resistances (R1, R2 and R3) and unknown resistance Rx.

The Wheatstone bridge is said to be balanced when no current flows through the galvanometer

Wheatstone Bridge

Hence

then

Therefore

Since thus

Dividing gives

1CBAC III 2DBAD III and

Potential at C = Potential at D

ADAC VV BDBC VV and

IRV

3211 RIRI and x221 RIRI

x2

32

21

11

RI

RI

RI

RI

3

1

2x R

R

RR

Example 18.17

A wheatstone bridge is used to make a precise

measurement of the resistance of a wire connector. If R = 1 kΩ & the bridge is balanced by adjusting P such that P =

2.5 Q, what is the value of X ?

Wheatstone Bridge:

Application (meter bridge)

The application of the Wheatstone bridge is Metre Bridge

The metre bridge is balanced when the jockey J is at such a

position on wire AB that there is no current through the

galvanometer. Thus the current I1 flows through the

resistance Rx and R but current I2 flows in the wire AB.

0

Accumulator

Jockey

Thick copper strip

(Unknown resistance)(resistance box)

Wire of uniform resistance

xR

A

ε

G

B

R

J

2l1lI I

1I

2I

1I

Wheatstone Bridge:

Application (meter bridge)

Let Vx : p.d. across Rx and V : p.d. across R,

At balance condition,

By applying Ohm’s law, thus

Dividing gives

AJx VV JBVV and

AJ2x1 RIRI JB21 RIRI and

A

ρlR 1

AJ JB2

AJ2

1

x1

RI

RI

RI

RI where and

A

ρlR 2

JB

A

ρl

A

ρl

R

R

2

1

x

Rl

lR

2

1x

Exercise 18.8(b)

The circuit shown in figure is

known as a Wheatstone

bridge. Determine the value

of the resistor R such that

the current through the 85.0

resistor is zero.

(Physics,3th edition, James

S. Walker, Q93, p.731)

ANS: 7.50

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Chapter 19:

Magnetic

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