resistor and resistance shatin tsung tsin secondary school mr. c.k. yu
TRANSCRIPT
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?