Chapter 15 Electric Circuits

Chapter 15 Electric Circuits
Physics Beyond 2000 Chapter 15 Electric Circuits

Electric current An electric current is a flow of electric charges through a conductor. Current = rate of flow of charges Constant current Changing current unit: ampere A

Definition of ampere and coulomb
Definition of 1 Ampere of current: Two infinitely long parallel wires 1 m apart 3. The forces on the wires are 2 × 10-7N m-1 F F 1m 4. The current in each wire is 1 ampere 2. The same amount of currents are in the wire I = 1A I = 1A

Definition of ampere and coulomb
Definition of 1 coulomb (C): Q = I.t 1 C = The amount of charges flowing through in 1 second with a current of 1 A. I = 1A t = 1 s

Charge-carriers The charge-carriers move and form the current.
The type of charges carried by a charge-carrier may either be negative or positive depending on the material.

Charge-carriers Substance charge-carrier Type of charge Metal
Free electrons Negative charge Electrolyte Positive ions Positive charges negative ions Negative charges Gas discharge tube Thermionic electrons

Charge-carriers Charge-carriers in electrolyte: positive and negative ions. + - positive ion negative ion negative electrode positive electrode water

Electric field in a circuit
The switch is open.The p.d. of AB is zero. Not any current to flow. 1. Cell with emf ξ 2. Switch S A B 3. Resistor R

Close the switch S. An electric field is established inside the circuit. ξ S R A B E

The electric field exerts force on the charge-carriers (electrons). Charge-carriers are accelerated. ξ S R A B E

The charge-carriers are also retarded by the collision with atoms. ξ S A B R E

As a result, the charge-carriers are moving at constant speed. ξ S R A B E

Drift velocity Charge-carriers in a metal move with constant speed under an electric field. The constant speed is called the drift velocity vD. E = electric field strength - vD

Drift velocity Consider a conductor with n charge-carriers per unit volume. E = electric field strength - vD

Drift velocity In time Δt , all the charge-carriers in the shaded region have passed the imaginary cross-section. cross-sectional area A vD. Δt E = electric field strength - vD

Drift velocity What is the number of charge-carriers ΔN in the shaded area? cross-sectional area A ΔN = nA.(vD.Δt) vD. Δt E = electric field strength - vD

Drift velocity Suppose each charge-carrier has charge q. What is the amount of charges ΔQ passing the cross-sectional area in time Δt? cross-sectional area A ΔQ = q.ΔN = q.nA.(vD. Δt) vD. Δt E = electric field strength - vD

Drift velocity What is the current I in the conductor? cross-sectional
area A vD. Δt electric field strength - vD

Drift velocity So the drift velocity of the charge-carriers in a conductor is cross-sectional area A vD. Δt electric field strength - vD

Drift velocity In small cross-sectional area, charge-carriers
So the drift velocity of the conductor is In small cross-sectional area, charge-carriers would move faster if other quantities are the same. In semi-conductor, n is small. The drift velocity is fast.

Drift velocity In metal, the charge-carriers are electrons
So the drift velocity of the conductor is In metal, the charge-carriers are electrons which are in random motion at very fast speed. This random speed does not contribute to the current.

Example 1 Drift velocity of the charge-carriers in a copper wire. (10-5 ms-1)

Signal and drift velocity
The drift velocity of electrons in a circuit is very small. The electric signal travels at a very fast speed in the circuit. It is close to the speed of light.

Electromotive force 1. Suppose that the charge-carriers are positively charged. 2. The charge-carriers gain energy from the battery when they flow from the low potential to high potential inside the battery. 4. The charge-carriers lose all energy when they return to the battery. 3. The charge-carriers lose energy when they move from high potential to low potential in the circuit.

Electromotive force 1. Suppose that the charge-carriers are positively charged. 2. The battery does work on the charge-carriers in the battery. The energy transferred to one coulomb of charge is called the electromotive force of the battery.

Electromotive force The electromotive force ξof a source
is the energy transferred into electrical energy per unit charge within the source. unit: J C-1 or V, volt where Q is the amount of charge through the battery and U is the total energy provided by the battery to the charge Q.

Electromotive force The electromotive force ξof a source
exists though there is not any current.

Example 2 The e.m.f. (electromotive force) of an AA cell.
Ah (ampere-hour) is a unit for charge. Q = I.t

Combination of cells Cells in series Cells in parallel

Cells in series ξ= ξ1 + ξ2 +ξ3 Produce a high e.m.f. ξ ξ1 ξ2 ξ3

Cells in parallel Total e.m.f. = ξ It can supply a larger current.
It lasts longer. It decreases the internal resistance of the source. ξ ξ

Potential difference (p.d.)
The p.d. between two points in the circuit is the amount of energy change as one coulomb of charge passes from one point to another. Unit: V (volt) An amount of charge Q passes from X to Y 2. The charge loses electrical energy U X Y 3.

Voltage Voltage is a general term.
It may refer to the e.m.f. of a source or the p.d. between two points. Unit: V (volt)

Measurement of voltage
Use voltmeter Use CRO to voltmeter or CRO to voltmeter or CRO I X Y

Example 3 First find the amount of charge of one mole of electrons.
U = QV

Voltage and electric field strength
A p.d. V between points X and Y. d 2. The separation XY is d. X Y 4. The average electric field strength is E = V/d. 3. There is an electric field E between X and Y.

Example 4 Use E = V/d and F = e.E E d
Note that the electrons move with uniform drift speed.

Resistance The p.d. between the ends of the conductor is V V I
2. A current I passes through the conductor 3. The resistance of the conductor is R = V/I 4. Unit of R is ohm, Ω

I-V graph Pure metal R = constant or V  I (Ohm’s law) The slope gives
V

I-V graph Bulb filament.
The filament resistance R increases with temperature. Ohm’s law is not obeyed. I V

I-V graph Semi-conducting diode
Current increases sharply in forward bias. Current is zero in reverse bias. I V current starts to increase

Dynamic resistance Rd Find the slope from the I-V graph Use I V

Dependence of resistance
Depending on the physical dimension such as length ( ) and cross-sectional area (A). Depending on the material which is described by its resistivity (ρ). A

Resistivity Material Resistivity ρ (Ωm) silver 1.62 × 10-8 copper
1.69 × 10-8 tungsten 5.25 × 10-8 pure silicon 2.5 × 103

Effect of temperature (metals)
Resistance increases uniformly with temperature for metals. θ/oC resistivity ρ ρo ρ= ρo (1 + α.θ) αis called the temperature coefficient.

Effect of temperature (metals)
An increase in temperature increases the thermal agitation of atoms. The interaction between atoms and electrons increases. The resistance increases. atom electron

Superconductivity Some metals or alloys lose the resistance at very low temperature. No dissipation of energy when current flows in a superconductor.

Internal resistance of battery
When current flows inside a battery, energy is also lost. The battery has internal resistance. The e.m.f. of a battery > The terminal voltage of a battery. e.m.f. describes the provision of power  = e.m.f. r 2. terminal voltage describes the loss of power V = terminal voltage

To measure the internal resistance r of a battery
2. An ammeter to measure the current I. 3. A variable resistor R to adjust the current R A I I  r A battery with emf  and internal resistance r

Express R in terms of , I and r. A  r R I

Plot the graph of R- How to find r from the graph? A  r R I R -r

The internal resistance r of a battery
We assume that the internal resistance is a constant. A  r R I R -r

Combination of resistors
Resistors in series Resistors in parallel

Resistors in series A common current passing through all resistors The equivalent resistance R of the system is R = R1 + R2 + R3 R1 R2 R3 I

Resistors in parallel Same voltage across each resistor The equivalent resistor R gives R1 R2 R3 V

Power and heating effect
When a current I passes through a resistor of resistance R and the p.d. is V, some electric energy is converted into internal energy, the power output is I R V

Power of a cell If the e.m.f. of a cell is ξand it provides a current I for an external circuit, then the power of the cell is  r R I

Example 5 Note that the internal resistor r is not counted in the power output. V power output  r R I

Power output and external resistance
The power output Pout is power output power output Pmax R r  r R I

The power output is maximum when R = r. power output power output Pmax R r  r R I

What is the efficiency when R = r? Hint: efficiency = Pout/Po power output Efficiency  r R I 100% 50% R r

Example 6 Use a cell of r = 2 Ω

Power transmission High voltage is used for transmission of electric power.

Power transmission What is the loss of power in the cable? (Use IL, the current in the cable, and R, the resistance of the cable.)

Domestic Electricity Live wire: with potential
changing between positive and negative 3. Electrical devices are connected in parallel. Same p.d. for each device. mains 2. Neutral wire: maintaining at zero potential by earthing in the power station.

Ring circuit live neutral A B C 1. Same p.d. for each device.

Ring circuit live neutral 1. Same p.d. for each device. A C
2. If a part is broken, can it be operated? B YES, it can.

Combination of two cells in parallel with same e.m.f.
A cell with emf  and internal resistance r1 3. Connect them in parallel 2. A cell with emf  and internal resistance r2 4. Note that there is not any current flow between the cells because their emf are the same.

Combination of two cells in parallel with different e.m.f.
A cell with emf 1 and internal resistance r1 5. There is current flowing from the cell of high emf to that of low emf 2. A cell with emf 2 and internal resistance r2 4. Connect them in parallel 3. 1 > 2

Example 6 (p.327) A cell with emf 14V and internal resistance 5.
5. There is current flowing from the cell of high emf to that of low emf. Find the current. 2. A cell with emf 12V and internal resistance 15 . 4. Connect them in parallel 3. 1(14V) > 2(12V)

Combination of two cells in parallel with same emf
 r1 Two cells of emf  are connected in parallel. 2. Both cells provide current I for the external resistor R. I1 I2 I  r2 3. Set up equations for these quantities. R

 r1 I1 + I2 = I (1) I1 I2  r2 I I R

 r1 I1 + I2 = I (1) I1  = I.R + I1.r (2) I I R

I1 + I2 = I (1)  = I.R + I1.r (2) I2  = I.R + I2.r (3)  r2 I I R

Example 6 Given:  = 1.5V, r1 = 2 , r2 = 4  and R = 6 .  r1 I1 I2
Find the currents and the power output. I I R

I-V characteristics Linear: Obey Ohm’s law. Ohmic device.
Non-linear: Not obey Ohm’s law. Non-ohmic device. I Non-ohmic Ohmic I V V Note that R =

I-V characteristics I R I Ohmic V V = 3 V r = 1  Find R if I = 1 A.

I-V characteristics Junction diode I/mA I + - 20 VD VD/V 0.8
VD is 0.8 V for a current of 20 mA. If the current is too large, the diode would be damaged.

I-V characteristics Junction diode I/mA I R 0.8V 20 VD/V 0.8
= 3V, r = 0 Find the suitable value of R. What would happen if R is too big or too small?

I-V characteristics Thermionic diode I/mA I 2 VD 4 VD/V
2 4 I VD The maximum current is 2.0 mA. This current is called the saturation current.

I-V characteristics Thermionic diode I = 2 mA I/mA R 2 VD 4 VD/V
2 4 R Find the minimum value for R. =12V, r = 0

Kirchhoff’s rules 1. Kirchhoff's Junction Rule,
2. Kirchhoff's Loop Rule and 3. Ohm's Law (i.e. V/I = R, where I is the current, V is the voltage, and R is the resistance).

Kirchhoff’s rules 1. Kirchhoff's Junction Rule
The algebraic sum of the currents at any branch point or junction in a circuit is zero. I1 I2 I3 I4 I5 Note that it is necessary to add a minus sign before I4 and I5 because they are leaving the junction. I1 + I2 + I3 - I4 - I5 = 0

Kirchhoff’s rules 2. Kirchhoff's Loop Rule
The algebraic sum of the potential changes around any complete loop in the network is zero. R3 C D Note the the potential drops by IR when current passing a resistor. It is necessary to add a minus before I.R following the current. I3 I2 B E R2 2 R4 I1 A F R1 1

Kirchhoff’s rules 2. Kirchhoff's Loop Rule
The algebraic sum of the potential changes around any complete loop in the network is zero. R3 C D Consider the loop ABEF: I3 I2 1-I1R1-I2R2 - 2 -I2R4 = 0 B E R2 2 R4 I1 A F R1 1

Kirchhoff’s rules circuit 1 circuit 2 circuit 3 circuit 4 circuit 5

Bridge circuits 4 resistors are connected as a bridge circuit. Two currents in the bridge circuit. R1 R2 R3 R4 I I1 I2 B A C D I = I1 + I2

Bridge circuits I = I1 + I2 If the potentials at B and C are equal,
Connect BD with a galvanometer. Is there any current passing the galvanometer? R1 R2 R3 R4 I I1 I2 A B D C I = I1 + I2

Example 7 The galvanometer shows no reflection 
No current flows between B and D The potentials at B and D are equal

Measuring devices Potentiometer

Measuring devices Potentiometer
To measure small p.d. To measure the internal resistance of a cell To calibrate a voltmeter To measure resistance Advantage: It can measure the voltage of a device without drawing any current from the device.

Potentiometer An accumulator provides a steady current for a long period of time. A resistance wire with uniform cross-section. 1. accumulator Vo (2V) I A B + - 2. A uniform resistance wire of length

To measure an unknown voltage (p.d.)
I 5. Both + ends are connected 4. A slide to contact wire AB. A B + - + - 3. A protective resistor V Device with p.d. 2. A galvanometer to detect current through the device

To measure an unknown voltage (p.d.)
B I + - V P Vo Move the slide on AB to find a balance point. 2. The galvanometer does not deflect when P is the balance point. 3.Show that

Using the protective resistor
1.Before the balance point is found, the current passing the galvanometer is large. Use the protective resistor to decrease the current. A B I + - V P Vo 2. When the balance point is near, the current passing the galvanometer is small. Short the protective resistor to increase the current.

Calibrating a potentiometer
Use a standard cell (e.g. Weston cell with emf = V at 20oC) to calibrate a potentiometer. A B I + - P Vo standard cell

Find the balance point P and measure the length AP. Find the voltage per unit length  of the resistance wire. A B I + - P Vo standard cell What is the voltage across AP? 2. So what is ? 3.  =

Use this calibrated potentiometer to measure the p.d. V of a device. A B I + - P’ Vo device V 1. V = .AP’

No balance point is found
The terminals of the device are reversed. The p.d. of the device is too large. V > Vo. A B I + - P’ Vo device V

Measure the emf of a cell
Find the balance point P emf V = AP   where  is the voltage per unit length. 3. Why is the result the emf? Why not the p.d.? A B I + - P Vo cell with emf V It is because there is not any current through the cell. The pd is equal to the emf.

Example 8 The SI unit of  is V m-1.

Measuring very small p.d.
AP would be too short if the measured V is very small because V =  .AP. The error of measurement is too big. A B I + - P Vo device with very small V

Measuring very small p.d.
Add a variable resistor R in series with the resistance wire. Adjust R so that VAB is just greater than V. Must re-calibrate the potentiometer when R is changed. A B I + - P Vo device with very small V R

Examples 9-10 The emf of a thermocouple is very small.
Adjust R so that the balance point P is near B. A B I + - P Vo R cold junction hot junction V

Measure the internal resistance r of a cell
B I + - P1 Vo S R r ‘ 1. The cell 2. r is the internal resistance 3. R is a known resistor 4. S is a switch

With switch S open, locate the balance point P1. 2. Measure length AP1. 3. The length AP1 represents the emf ’ of the cell. A B I + - P1 Vo S R r ‘ 4. There is not any current flowing through the cell

With switch S closed, locate the balance point P2. 2. Measure length AP2. 3. There is current I2 passing the cell and the resistor R. 4. Length AP2 represents the voltage V across R. Measure the internal resistance r of a cell A B I + - P2 Vo S R r ‘ I2 V

1. Note that the two currents I and I2 do not mix up. A B I + - P2 Vo S R r ‘ I2 V 2. We have ’ = I2.R + I2.r

‘ =   AP (1) V =   AP (2) ’ = I2.R + I2.r (3) V = I2.R (4) ‘ r I2 R V Find r in terms of AP1, AP2 and R.

Example 11 First find the voltage per unit length .
The terminal voltage V is the voltage across the cell with current flowing. The e.m.f. is the voltage across the cell without current flowing. The above two are not equal if the cell has internal resistance r.

Calibrating a voltmeter
To check that the reading of a voltmeter is correct. V + -

1. V =  . AP A P B + - V R R’ E

+ - P Vo V R R’ E 1. V =  . AP 2. Connect a voltmeter to measure the voltage across R. Compare its reading with the above value 3. Adjust R for other readings of the voltage. Each time locate the new balance point

Measure resistance R A B I + - P Vo R E R’ Measure V across R using the potentiometer. V= ξ.AP Read the current I through the resistor. R =

Electrical meters Ideal measuring instruments: should not change the system being measured. Ammeter: its resistance should be as small as possible. Voltmeter: its resistance should be as large as possible.

Moving-coil galvanometer
It can measure a small current.

Centre-zero galvanometer

Can be changed into an ammeter or voltmeter. Typical data of a moving-coil galvanometer: f.s.d. current = 1.0 mA f.s.d. voltage = 0.15 V resistance = 150 Ω + -

Ammeter It is connected in series in the circuit. R A A I
Note that the total resistance of the circuit is R + r where r is the resistance of the ammeter.

Adapt a galvanometer to measure a large current (an ammeter)
Shunt: add a resistor in parallel with the galvanometer. The adapted galvanometer has a smaller resistance. moving-coil galvanometer the ammeter + - shunt R

Adapt a galvanometer to measure a large current (an ammeter)
Example: A moving-coil galvanometer with f.s.d. current 1.0 mA is adapted to measure 1.0A at f.s.d.. What is the resistance of the shunt? moving-coil galvanometer I = 1.0A shunt I2 I1=1.0mA V R = 0.15 R Note that the p.d. across R = the p.d. across the galvanometer.

voltmeter It is connected in parallel in the circuit. V R V I
Note that the total resistance of the circuit is R.r/(R + r) where r is the resistance of the voltmeter.

Adapt a galvanometer to measure a large voltage (a voltmeter)
Multiplier: add a resistor in series with the galvanometer. The adapted galvanometer has a larger resistance. multiplier R + - voltmeter

Adapt a galvanometer to measure a large voltage (a voltmeter)
Example: A moving-coil galvanometer with f.s.d. voltage 0.1 V is adapted to measure 15 V at f.s.d.. What is the resistance of the multiplier? R 15V 0.1V I=1.0mA I R = 15 k Note that the same current is passing through the multiplier and the galvanometer.

Load effect of a voltmeter: a voltmeter draws current
1. Without voltmeter 2. With a voltmeter of resistance 15 k 6V 5 k I = 0.6 mA 3V 6V 5 k I = mA V 15k 3.43V 2.57V 0.171 mA 0.515 mA

Load effect of a voltmeter: a voltmeter draws current
The best solution is to add a voltage follower. A voltage follower draws a current as small as 0.1 nA from the circuit. 6V 5 k V 0.6mA < 0.1nA - + The voltmeter draws current from the voltage follower. Not from the circuit. 15k

Ohm-meter (digital) To read the resistance directly from the ohm-meter.

Ohm-meter (analog) cell with emf  2. Moving-coil galvanometer
3. Variable resistance for setting zero Z + - 4. Connect resistor to these terminals for measurement

Ohm-meter (analog) 1. When measuring a resistor R, + - R  I Z
2. The deflection of the galvanometer is proportional to I. 3. Adjust the scale of the galvanometer to read R from the deflection. 4. The scale is non-linear.

Measuring resistance by voltmeter-ammeter method
Set up either of the following circuits. Measure V and I. Calculate R from 1. For small R 2. For big R A V R A V R I I

Measuring resistance by voltmeter-ammeter method
Discuss why there are two different circuits. Assume values for R in your discussion. 1. For small R 2. For big R A V R A V R I I

Multimeter A-V-O meter: ammeter, voltmeter and ohm-meter. (Analog)
(Analog) (Digital)

Multimeter moving-coil galvanometer shunts To set zero Z multipliers E
100mA 1V E 10V 100V  + _ unknown resistor

Shunts and currents Given: f.s.d. current of the galvanometer = 1.0 mA
resistance of the galvanometer = 150  resistance of each shunt = 0.15  Find: the f.s.d current in each of the following adapted galvanometer. I1 I2  1A  0.5A R R R Resistance of the shunt is smaller. Resistance of the shunt is larger.

Multimeter To measure large current shunts E Z + _ To set zero I
100mA + _ To set zero To measure large current I 1. external current flows into the multimeter 2. external current through the shunt (one resistor, bigger current). 3. external current through the galvanometer 4. external current flows out of the multimeter

Multimeter To measure small current 3. external current through
the galvanometer To measure small current shunts 2. external current through the shunt (three resistors, less current). To set zero Z 1A 10A 100mA E + _ 4. external current flows out of the multimeter 1. external current flows into the multimeter I I

Multiplier and voltages
Given: f.s.d. current of the galvanometer = 1.0mA f.s.d. voltage of the galvanometer = 0.15V resistance of the galvanometer = 150  resistance of each multiplier = 15 k Find: the f.s.d voltage in each of the following adapted galvanometer. I1 Resistance of the multiplier is smaller. R V1 R V2 I2 Resistance of the multiplier is larger. 15V 30V

Multimeter To measure a high voltage 3. external current through
the galvanometer 2. external current through the multiplier (three resistors, higher voltage). To set zero Z multipliers 1V E 10V 100V 4. external current flows out of the multimeter _ + 1. external current flows into the multimeter V

Multimeter To measure a low voltage 3. external current through
the galvanometer 2. external current through the multiplier (one resistor, lower voltage). To set zero Z multipliers 1V E 10V 100V 4. external current flows out of the multimeter _ + 1. external current flows into the multimeter V

Multimeter 4. The galvanometer deflects to show the resistance R.
To measure R short the inputs and set zero with Z. shunts To set zero Z multipliers I E 3. The internal cell provides the current.  + _ 2. connect an unknown resistor R to the inputs 5. The current depends on R. R

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