Resistor and ResistanceResistor and Resistance

Shatin Tsung Tsin Secondary SchoolShatin Tsung Tsin Secondary School

Mr. C.K. YuMr. C.K. Yu

How do electrons (charges) flow within a conductor

Direction of electron flows

Direction of currentCurrent : flow of positive charges

How do electrons (charges) flow within a conductor

When electrons travel through, they bump When electrons travel through, they bump onto the outer electrons or the atoms and onto the outer electrons or the atoms and energies are given to the conductor.energies are given to the conductor.

Ohm’s Law

(i) Conductors (導電體 ) are material which can conduct electricity (傳電 ).

(ii) Electrons (negative charges) or positive charges can go through conductor.

(iii) Negative charges give energy to the conductor when they travel through the conductor.

Ohm’s Law

(iv) If there is a potential difference across a conductor, a current will flow through the conductor.

(v) Ohm discovered (發現 ) that, in somesome conductors, the current (I) and the potential difference or voltage (V) across the conductor is directly proportional.

Ohm’s Law(vi) So, his discovery is called Ohm's law.

Voltage / V

Current / A

The proportional The proportional constant (the slope) is constant (the slope) is called the called the resistanceresistance of of that materialthat material

Ohm’s Law-summary

Ohm’s law states that the p.d. across an ohmic conductor is directlydirectly proportionalproportional to the current through it, provided that the temptemperatureerature and other physical conditions are constant (the same). The proportional constant is called resistance resistance ..

A material is called ohmic material if it follows the ohm’s law

Experiment – Ohm’s Law

Objective : To verify the ohm’s law

Apparatus:

1 power supply1 resistor (red)

1 voltmeter 1 ammeter

1 circuit board connecting wires

1 switch

Experiment – Ohm’s Law

• Your power supply consists of 4 electric cells. In this experiment, you are to measure the relationship between the currentcurrent passing through and the potentialpotential differencedifference across a resistor.

Experiment – Ohm’s Law

1. Connect a red resistor, an ammeter, a switch and a power supply (4 cells) in series as shown in Figure 1. Figure 1

Experiment – Ohm’s Law

2. Close the switch and measure the current shown in the ammeter, record the reading in Table 1.

Table 1

No. Reading of Ammeter (I/A)

Reading of Voltmeter (p.d./V)

Remark

1 1 cell

2 2 cells

3 3 cells

4 4 cells==== ================

Experiment – Ohm’s Law

3. Keep the switch closed. Use the voltmeter to measure the potential difference across the red resistor as shown in figure 2. Record the reading in Table 1.

Figure 2

Experiment – Ohm’s Law

4. Repeat steps 1 to 3 with different numbers of electric cells in the power supply. Also record the results in Table 1.

Experiment – Ohm’s Law5.In the graph below , plot the graph of V against I.

V/V

I/A

Experiment – ConclusionThe proportional constant (resistance) of the

resistor is : ___________ ΩOhm’s law states that the p.d. across a

conductor is __________ proportional to the current through it, provided that the ___________ and other physical conditions are constant.

The ratio of p.d./current is the resistance of the conductor. It is measured in ___________.

directly

temperature

Ohm, or Ω

HOT questions

If the battery of a circuit is exhausted, what will happen to the current in the circuit?

• __________________________________

What are the possible reasons if the current in a circuit increases?

• ____________________________________________________________________

The current will decrease and become zero.

Either the resistance is lower or the supplysource is of higher emf.

Resistors

(i)Any conductors which work under Ohm's Law are called resistor.

(ii)The proportional constantproportional constant (正比常數 ) of voltage and current is called the resistanceresistance (電阻值 ) of that conductor.

(iii)Unit of resistance is "Ohm", or .

Resistors (iv)The symbol of a resistor in a circuit :

(v)The mathematical equation for Ohm’s law is:

Where R : the resistance of the resistor, V : the voltage across the resistor,

I : the current flow through the resistor.

R

R = V

I

12e.g.

(c) Simple resistor circuit3V

15

0.2 A

R = 15

V = 3 V

I = 0.2 A

The values of R, V, I follows the ohm’s law.The values of R, V, I follows the ohm’s law.

15 Ω

Example

5 C of charge passes a resistor in a circuit in 2 s, the total electrical energy dissipated by the charge is 8 J.

(i) What is the resistance of the resistor ?V = E / Q = 8 J / 5C = 1.6 V

I = Q / t = 5 C /2 s = 2.5 AR = V / I = 1.6 V / 2.5 A = 0.64

Example

5 C of charge passes a resistor in a circuit in 2 s, the total electrical energy dissipated by the charge is 8 J.

(ii) What is the power dissipation of the resistor ?

P = E / t = 8 J / 2 s = 4 W

Revisit of Electric Power

P =E

t( by definition)

P = V x I (V=E/Q,I=Q/t, V x I=E/ t)

P = V2 / R (I = V/R)

P = I2 R (V = I R)

Summary TableQuantities Symbol Unit Unit symbol

Time t Second s

Charge Q Coulomb C

Current I Ampere A

Electromotive force e.m.f Volt V

Voltage V Volt V

Potential difference p.d. Volt V

Power P Watt W

Resistance R Ohm

Resistors in series

(i) Resistors can be connected together at each end.

(ii) One of the basic methods is to connect the resistors one by one in a line, called in series.

Resistors in series

(iii) The following diagram shows three resistors connected in series :

(iv) they can be considered as a single they can be considered as a single resistorresistor, with a total resistance RT.

R R R1 2 3

R R R1 2 3 RT

RRTT follows Ohm’s law follows Ohm’s lawAll charges flow through RAll charges flow through R11 will also flow will also flow

through Rthrough R22 and R and R33

Resistors in series

In calculation : RRTT = R = R11 + R + R22 + R + R33 + ….. + …..

Example :

R1 = 5, R2 = 3 , R3 = 5

If the above three resistors are connected in series, the total resistance is

______________________

5 5 + 3 + 3 + 5 + 5 = 13 = 13

Resistors in Parallel(i) Another method is to connect the

resistors one over another at two ends, called in parallel.

(ii) The diagram shows three resistors connected in parallel :

Charges flowing from A to B will only go through one of the resistors in parallel.

R1

R2

R3

A B

Resistors in Parallel

(iii) These resistors can be considered as a single resistor, with a total resistance RT

R1

R2

R3

Resistors in Parallel

(iii) These resistors can be considered as a single resistor, with a total resistance RT

R1

R2

R3

RT

RRTT follows Ohm’s law follows Ohm’s law

Resistors in ParallelIn calculation :

321

1111

RRRRT + … … …

Example : R1 = 5 , R2 = 10 , R3 = 5

If the above three resistors are connected in parallel, the total resistance is __ .22

Examples

• Two resistors of 5 ohms and 15 ohms are connected in series. The total resistance is ______ ohms.2020

Examples

• Three resistors of the same resistance are connected in series and the total resistance is 15 ohms. What is the resistance of each resistor? 5 ohms5 ohms

Examples

• Two resistors of the same resistance are connected in parallel and the total resistance is 5 ohms. What is the resistance of each resistor? 10 ohms10 ohms

Examples

• Two resistors are connected in parallel and the total resistance is 4 ohms. One of them is 5 ohms, what is the resistance of the other resistor? 20 ohms20 ohms

Examples

• Three resistors are connected in parallel and the total resistance is 2 ohms. Two of them are 5 ohms and 10 ohms, what is the resistance of the third resistor? 5 ohms5 ohms

Class Example

(i) If 2 resistors of 2 Ω and 5 Ω are connected in series, the total resistance is :

R1 = 2 , R2 = 5 By RT = R1 + R2

= 2 + 5 = 7 Ω

Class practices(ii) If 4 resistors of 2, and 9 , are connected in series, the total resistance is :

The total resistance is : 2+3+8+9 = 22

Class practices

(iii) If 2 resistors of 2 and 5, are connected in parallel, the total resistance is :

The total resistance : 1/R = 1/2+1/5 = 7/10 R = 10/7 or 1 3/7

Class practices

(iv) If 4 resistors of 2, , 8and 9, are connected in parallel, the total resistance is :

The total resistance : 1/R = 1/2+1/3+1/8+1/9

= (36+24+9+8)/72=77/72

R = 72/77

Class Practices

If 3 resistors of 2, 3, 8 are connected in series and then in parallel with a resistor of 9, the total resistance of this connection is :

The total resistance : 1/R = 1/(2+3+8)+1/9

= (9+13)/117=22/117

R = 117/22 or 5 7/22

HOT – Home Practice

• Only resistors of 5 ohms are given. How do we connect the resistors to form a total resistance of 4 ohms?

SummarySummary

(i) Resistors connected either in series or in parallel can be considered as a single resistor of resistance RT.

(ii) The current and the voltage across the resistors or the equivalent resistor can also be calculated using the equations :

RT = VT / IT

Simple measurement experiments

• Objectives:

– Measure current and voltage with ammeter and voltmeter

– Find out the characteristics of resistors connected in parallel

Construct the circuit as shown by connecting a 6V battery and two identical resistors, R1 and R2 , in parallel. Show your connection to your teacher. 6V

I R1

R2

I1

I2

I

6V

I R 1

R 2

I 1

I 2

I

A

6V

I R 1

R 2

I 1

I 2

I

A

I1=_______ A I2=_______ A

Connect an Ammeter in series with each resistor to measure current I1 and I2.(I1 and I2 can be called branch currents)

Then connect an ammeter in series with the battery as shown in the diagram. Measure the current I. 6V

I R 1

R 2

I 1

I 2

I

A

I=_______ A

Compare the three reading, I1, I2, and I I=_______ + ________ = ________ AI1 I2

Finally, connect a Voltmeter, in turns, in parallel across each resistor to measure p.d. across each of them respectively

6V

V

V

p.d. across R1 =____V,

p.d. across R2 =____V.

Conclusion : The p.d. across each parallel branch is __________.the same

Summary of resistors Summary of resistors in series or in parallelin series or in parallel

Resistors in parallel :

Voltages across all branches in parallel are the same. V = VV = V1 1 = V= V2 2 = V= V3 3 =….=….

The current before and after the resistors in parallel is the sum of the currents of all branch. I = II = I11+I+I22+I+I33+ ….+ ….

Summary of resistors Summary of resistors in series or in parallelin series or in parallel

Resistors in Series :

Current of each resistor in a series is the same. I = II = I1 1 = I= I2 2 = I= I3 3 =….=….

The p.d. (voltage) across the beginning and the end of the resistors in series is the sum of the p.d. across each resistors.

V = VV = V11+V+V22+V+V33+ ….+ ….

HOT Questions

Which of the following is true in the following description about a wire.

The longer the wire, the higher the resistance of the wire.

The larger the diameter of the wire, the higher the resistance of the wire

Comparison table

In series In parallel

Current I = I1=I2=I3=… IT=I1+I2+I3+…

Voltage, p.d. VT=V1+V2+V3+… V=V1=V2=V3+…

Total Resistance

RT=R1+R2+R3+… ...1111

321

RRRRT

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(1)the total resistance of the circuit,(2)the current, I of the circuit,(3)the current I1 passed through

the resistor R1,

(4)the current I2 passed through the resistor R2,

(5)V1 , the potential difference across the resistor R1, and

(6)V2 , the potential difference across the resistor R2.

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(1)the total resistance of the circuit,

RT = 15 + 9 = 24

RRT = R = R1 + R + R2 + R + R3 + ..+ ..

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(2)the current, I of the circuit,

R = V/ IR = V/ I2424 = 6V / I = 6V / I I = 0.25 AI = 0.25 A

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(3)the current I1 passed through

the resistor R1,

I = II = I11=I=I22

II11 = 0.25A = 0.25A

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(4)the current I2 passed through

the resistor R2,

I = II = I11=I=I22

II22 = 0.25A = 0.25A

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(5)V1 , the potential difference

across the resistor R1, and

By By RR11 = V = V11/I/I11

1515 = V = V11 / 0.25 A / 0.25 A VV11 = 3.75 V

Advanced Example(i) Below shows a series circuit, calculate

6V

15

IR1 R2

9I2I1

V1 V2

(6)V2 , the potential difference

across the resistor R2.By RBy R22 = V = V22/I/I22

99 = V = V22 / 0.25 A / 0.25 A VV22 =2.25 V

Advanced Example(ii) Below shows a parallel circuit, calculate

(1)the total resistance of the circuit,(2)the current, I, passed through the circuit,(3)the p.d. across the resistor R1, (4) the p.d. across the resistor R2.(5)the current I11 passed through the resistor, and(6)the current I2 passed through the resistor R2 .

6V

15

IR1

R2

9

I1

I2

I

`(ii) Below shows a parallel circuit, calculate

(1)the total resistance of the circuit,

6V

15

IR1

R2

9

I1

I2

I

By By

1/R1/RTT = 1/15 + 1/9 = 8/45 = 1/15 + 1/9 = 8/45

RRTT = 5.625 = 5.625

...1111

321

RRRRT

Exact valueExact value

Advanced Example(ii) Below shows a parallel circuit, calculate

(2)the current, I, passed through the circuit,

6V

15

IR1

R2

9

I1

I2

IBy By RRTT = V = VTT/I/ITT

5.625 5.625 = 6V / I = 6V / II = 1.067 AI = 1.067 AI = 1.07 A I = 1.07 A

Advanced Example(ii) Below shows a parallel circuit, calculate

(3)the p.d. across the resistor R1, 6V

15

IR1

R2

9

I1

I2

I VVTT = V = V11=V=V22=V=V33= ….= ….

Then p.d. = 6VThen p.d. = 6V

Advanced Example(i) Below shows a parallel circuit, calculate

(4) the p.d. across the resistor R2.

6V

15

IR1

R2

9

I1

I2

I VVTT = V = V11=V=V22=V=V33= ….= ….

Then p.d. = 6VThen p.d. = 6V

Advanced Example(i) Below shows a parallel circuit, calculate

(5)the current I11 passed through the resistor, and

6V

15

IR1

R2

9

I1

I2

I

By By R R11 = V = V11//II11

1515 = 6V / = 6V / I I11

II11 = 0.4 A = 0.4 A

Advanced Example(i) Below shows a parallel circuit, calculate

(6)the current I2 passed through the resistor R2 .

6V

15

IR1

R2

9

I1

I2

I

By By RR22 = V = V22//II22

99 = 6V / = 6V / I I22

II22 = 0.667 A = 0.667 A

Mixed Problem6V

12

IR1

R2

4

I1

I2

R3

9I3

I

Mixed Problem

1)The total resistance of the circuit

2)The current, I, passed through the circuit

3)The current I3 passed through the resistor R3

6V

12

IR1

R2

4

I1

I2

R3

9I3

I

4)The p.d. across the resistor R1

5)The p.d. across the resistor R2

6)The current I1 passed through the resistor R1 and7)The current I2 passed through the resistor R2

The total resistance of the circuit6V

12

IR1

R2

4

I1

I2

R3

9I3

I

The total resistance of the circuit6V

12

IR1

R2

4

I1

I2

R3

9I3

I

R = 1/(1/R1 + 1/R2)

6V

12

IR1

R2

4

I1

I2

R3

9I3

I

RT = 1/(1/R1 + 1/R2) + R3

RT = 1/(1/12 + 1/4) + 9= 3 + 9 = 12

1) The total resistance of the circuit

2) The current, I6V

12

IR1

R2

4

I1

I2

R3

9I3

I

RT = 1/(1/R1 + 1/R2) + R3

By By RRTT = V = VTT//IIT T , , 1212 = 6V / = 6V / II

II = 0.5 A = 0.5 A

3) The current, I3

6V

12

IR1

R2

4

I1

I2

R3

9I3

I

By By II33 = = II

II33 = 0.5 A = 0.5 A

The p.d. across R3

6V

12

IR1

R2

4

I1

I2

R3

9I3

I

By By RR33 = V = V33//II33

99 = V = V33 /0.5A /0.5AVV33 = 4.5 V = 4.5 V

4) The p.d. across R16V

12

IR1

R2

4

I1

I2

R3

9I3

I

R = 1/(1/R1 + 1/R2)

By By 6V = V + V6V = V + V3 3 , V = 6V - 4.5 V = 1.5 V, V = 6V - 4.5 V = 1.5 V

VV VV33

5) The p.d. across R26V

12

IR1

R2

4

I1

I2

R3

9I3

I

R = 1/(1/R1 + 1/R2)

The p.d. across RThe p.d. across R22 is also 1.5 V is also 1.5 V

VV VV33

6) The Current I16V

12

IR1

R2

4

I1

I2

R3

9I3

I

By By RR11 = V = V11//II11

1212 = 1.5V/ = 1.5V/II11

II11 = 0.125 A = 0.125 A

7) The Current I26V

12

IR1

R2

4

I1

I2

R3

9I3

I

By By RR22 = V = V22//II22

44 = 1.5V/ = 1.5V/II22

II22 = 0.375 A = 0.375 A

The EndThe End

Thank you. What should you do?