PHYSICS NOTES. W.R.David. Current Electricity

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1 PHYSICS NOTES W.R.David Current Electricity Syllabus:- Content Electric current Potential difference Resistance and resistivity Sources of electromotive force. Learning Outcomes Candidates should be able to: (a) show an understanding that electric current is the rate of flow of charged particles. (b) define charge and the coulomb. (c) recall and solve problems using the equation Q = It. (d) define potential difference and the volt. (e) recall and solve problems using V = W/Q. (f) recall and solve problems using P = V I, P = I 2 R. (g) define resistance and the ohm. (h) recall and solve problems using V = IR. (j) sketch the temperature characteristic of a thermistor. (Thermistors will be assumed to be of the negative temperature coefficient type.) (k) state Ohm s law (l) recall and solve problems using R = ρl/a. (m) define e.m.f. in terms of the energy transferred by a source in driving unit charge round a complete circuit. (n) distinguish between e.m.f. and p.d in terms of energy considerations. (o) show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power. Review of Basics Charge Charge is a property of certain particles. A particle with charge will experience a force in an electric field (or in a magnetic field if the charge is moving). Charge is measured in coulombs, C. The amount of charge on an object can be found using a coulomb meter. An electron always has a negative charge of coulombs. Protons have an equal amount of positive charge. 1

2 Quantisation of charge:charges are integer multiples of a the charge of an electron or proton.(± C).This means that the minimum charge possible is ± C. Mathematically this can be put as q = ±ne where q is the charge,n is an integer and e is ± C Using this equation we can find that, one coulomb is equal to the charge on electrons, which is a serious number of electrons(check this out). The Conservation of Charge:It is not possible to destroy or create charge.you can cancel out the effect of a charge on a body by adding an equal and opposite charge to it, but you can t destroy the charge itself. That s the Principle of the Conservation of Charge. Static Electricity:Static electricity is caused by the transfer of electrons from one object to another. Normally neutral atoms can lose or gain electrons to become either positively or negatively charged. These charged atoms are called ions.static electricity is never caused by the movement of protons.the easiest way to charge an object with static electricity is by using friction.(rubbing a glass rod with silk cloth charges the rod positive and the silk cloth negative) Current:Current is the rate of flow of charge; it is the amount of charge flowing per second through a conductor. The equation for calculating current is: I = q t Where: I = current (amps, A),q = charge flowing past a point in the circuit (coulombs, C),t = time taken for the charge to flow (seconds, s) hence a current of 1 amp is 1 coulomb of charge flowing past a point every second. Likewise a coulomb is the same as an ampere-second. (Note: if you plot a graph of current flowing against time, the area under the graph will equal the charge that has moved.) Current direction: Conventional current directionis taken from positive to negative.it should be noted that the electrons move in the opposite direction i.e from negative to positive. How to make the Charge to Flow?: First a conductor is required for the charge to flow through. Energy is required make charged particles move(this is usually provided by a battery,or electric power source) Work is being done on these charged particles to make them move,the voltage is a measure of the amount of energy that is provided per coulomb of charge.the equation for calculating voltage is: V = W q where W = amount of energy (joule, J),V = voltage (volt, V),Q = charge (coulomb, C) Hence 1 volt is also 1 Joule per Coloumb. Some key concepts:- 2

3 As the charged particles flow around a circuit they don t get used up; it is the energy that the charged particles carry that decreases as they move around the circuit. Note:(Imagine charges as runners going around the 200m track they run all the way round, but they lose energy as they run) So current is not used up - if 5 amps leaves the battery, there will be 5 amps in the circuit and 5 amps returning to the battery. Voltage changes as the charge moves around the circuit. The potential energy given to the charge is changed into heat energy in the circuit. An electron may leave a battery with 6 V, but will return to the battery with 0 V. This gives a change in potential of 6 V, hence the words potential difference. Measuring Current and Voltage: To measure current we use an ammeter. It is placed in series in a circuit to measure the amount of charge flowing through it per second. (You can compare it to a turnstile counting people into a stadium.) To measure voltage we use a voltmeter. It is placed in parallel to compare the potential at two different points, either side of a component. It can then measure the potential difference, or voltage across the component. Ohms Law: It states that physical conditions (like temperature, strain etc) remaining unchanged the current flowing through the conductor is directly proportional to the potential difference across it s ends. V I or V = IR R the constant of proportionality is called resistance. Unit of resistance is Ohm(Ω) 1Ohm(Ω) = 1 1volt(V ) 1Ampere Resistance:- Electrical Resistance is defined as the property of a material to oppose the flow of charges (current). When a battery is connected across a conductor,the charge carriers (in the case of metallic conductors the charge carriers are electrons) gradually move toward s one end of the conductor,in this process the charge carriers continually collide with other electrons and the atoms in the conductor as they gradually drift toward s the other end of the conductor.the passage of each electron is opposed by the conductors atoms and other electrons. In other words the passage of current is resisted by the conductor. Unit of Resistance: The unit of resistance is Ohm(Ω) Conductance:The inverse of resistance ( 1 ) is called conductance.the unit of conductance is R Siemens(S).Which is Ω 1 Key Points about Resistance Resistance varies with temperature Resistance is directly proportional to the length of the conductor and inversely proportional to the area of cross section of the conductor. 3

Circuit for verifying Ohms Law Figure 1: Ohm s Law circuit-note that the voltmeter is connected in parallel and the ammeter is connected in series Variation of V with I Figure 2: The variation of V

4 Circuit for verifying Ohms Law Figure 1: Ohm s Law circuit-note that the voltmeter is connected in parallel and the ammeter is connected in series Variation of V with I Figure 2: The variation of V with I:The inverse of the gradient gives the resistance Key features of Ohms Law: The resistance of a conductor changes with temperature-for example for metallic conductors the resistance increases with temperature which disturbs the linearity of the the V-I graph Figure 3: The dotted line represents the Ideal ohms law variation.the solid line shows the variation of Resistance due to heating causing the break down of linearity Ohms law is not universal there are components and materials which don t obey Ohms Law. These are called as non-ohmic conductors examples of such conductors are Diodes,semiconductors, thermistors etc.the graph bellow shows the variation of V with I for a diode (you will learn abut the diode later). 4

5 Figure 4: V-I graph for a Diode: notice the non linearity Temperature also affects the linearity of the V-I graph Thermistor:A thermistor is a special type of resistor which has been deliberately manufactured so that its resistance decreases as its temperature rises. A plot of its characteristics (a voltage against current graph)is shown below. Figure 5: The temperature dependence of Resistance for a thermistor:the resistance drops as temperature increases Uses of thermistors:thermistors are widely used as current limiters, temperature sensors, current protectors,etc. Temperature dependence of resistance for metallic conductors The resistance of metallic conductors increases with temperature as shown below: Figure 6: Resistivity ρ T of copper with Temperature 5

6 Resistivity(ρ): Factors affecting resistance: Length - The further electrons have to travel through material, the more collisions they will have so the higher the value of resistance.(r l) Area - a bigger area means that in any 1 second more electrons will be able to travel through a piece of wire. More electrons means more current which means less resistance.(r 1 A ) Material - Different materials have different resistances if we used a the copper wire instead of an iron wire of the same dimension there will be less current and a lot more resistance in the circuit. Resistivity(ρ) can be defined on the basis of these factors:- R l or R 1 A R = ρ l A Where ρ is the constant of proportionality and is called resistivity. Unit of resistivity:the unit of resistivity is Ohm m (Ωm) Conductivity(S): Conductivity is the reverse of resistivity. Unit of Conductivity:Conductivity is measured in Siemens/meter (Sm 1 ) S = 1 ρ = l RA Electrical Power and Energy As electricity is provides energy, there are a few energy and power equations that you need to be able to derive and use... Electrical potential is defined as the work done per unit charge or From the definition of current V = W q q = It Using the above two equations the workdone (W) by the electrical power source (say the battery) can be calculated as W = V It We know that: hence P ower(p ) = W orkdone time P ower(p ) = V It t = V I 6

7 So, Power = V I (Learn this!) Using Ohms law V = IR in the above equation we get P ower(p ) = I 2 R Similarly using I = V R in P = V I we get P = V 2 R The Chart below gives all the details about Electrical Formulea:- Internal Resistance:One of the main sources of electrical energy energy are cells and Batteries (group of Internally connected Cells),and all practical cells have internal resistance,the cause of the internal resistance is the collision of electrons (or ions) with the charge carriers within the electrolyte of the cell.due to internal resistance of the cell the cell loses energy in the form of heat (thats why cells become hot after some time). The bigger the internal resistance then the bigger the energy loss in the battery..internal Resistance can be represented as a resistance connected in series with the Battery. EMF (E) and Potential difference(v) The emf E is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell.its unit is volts The electromotive force (e.m.f) of a cell (source) as the work done by a source in driving one unit of charge around a complete circuit.note that emf is not a force (the name is a misnomer!) The potential difference (p.d) across a component in a circuit as the work done to drive a unit charge through the component. The total emf is equal to the sum of the p.d across all the components in a complete circuit- Including the drop across the internal resistance of the cell. The emf is always greater than the potential difference. 7

8 Figure 7: Measuring EMF Relation between emf internal resistance and Potential difference The circuit shown in figure 7 clearly shows the way emf is different from potential difference. Consider the circuit below the internal resistance is shown as resistance connected in series to the battery,the total emf is equal to the sum of the p.d across all the components in a complete circuit-including the drop across the internal resistance of the cell,using this concept we write an equation connecting the enf(e),internal resistance(r) and Potential difference(v ) Figure 8: Internal Resistance can be represented as a resistance(r) connected in series with the Battery but hence Effect of Internal resistance on Power Electrical power is given by The cell emf is given by E = IR + Ir V = IR E = V + Ir P = V I E = IR + Ir Where IR = V is the cell s potential difference. Multiplying the emf equation by I we get IE = I 2 R + I 2 r 8

9 The first term IE is the electrical produced per second (Power)by the cell The second term I 2 R is the Heat energy produced per second (Power) in the resistance R The third term I 2 r is the Heat energy produced per second (Power) in the internal resistance r Overall the equation tells that the power generated by the cell is equal to the power supplied to the resistor plus the power wasted inside the cell due to its internal resistance r. Some Questions from the past Papers END 1. Distinguish between the electromotive force (e.m.f.) of a cell and the potential difference (p.d.) across a resistor. 2. Fig. below shows the variation with applied potential difference V of the current I in an electrical component C. (a) (i) State, with a reason, whether the resistance of component C increases or decreases with increasing potential difference. (ii)determine the resistance of component C at a potential difference of 4.0 V. (b) Component C is connected in parallel with a resistor R of resistance 1500 and a battery of e.m.f. E and negligible internal resistance, as shown in Fig.below:- (i) On Fig. the graph, draw a line to show the variation with potential difference V of the current I in resistor R. (ii) Hence, or otherwise, use Fig. 6.1 to determine the current in the battery for an e.m.f. of 2.0V. 9

10 3. A household electric lamp is rated as 240 V, 60W. The filament of the lamp is made from tungsten and is a wire of constant radius m. The resistivity of tungsten at the normal operating temperature of the lamp is Ω. (a) For the lamp at its normal operating temperature, (i) calculate the current in the lamp, (ii) show that the resistance of the filament is 960Ω. (b)calculate the length of the filament. 4. A student set up the circuit shown in Fig The resistors are of resistance 15Ω and 45Ω. The battery is found to provide J of electrical energy when a charge of C passes through the ammeter in a time of s. (a) Determine (i) the electromotive force (e.m.f.) of the battery, (ii) the average current in the circuit. (b) During the time for which the charge is moving, J of energy is dissipated in the 45Ω resistor. (i) Determine the energy dissipated in the 15Ω resistor during the same time. (ii) Suggest why the total energy provided is greater than that dissipated in the two resistors. 5. (a) Define the resistance of a resistor. (b) In the circuit of Fig. 7.1, the battery has an e.m.f. of 3.00 V and an internal resistance r. R is a variable resistor. The resistance of the ammeter is negligible and the voltmeter has an infinite resistance. 10

11 The resistance of R is varied. Fig. 7.2 shows the variation of the power P dissipated in R with the potential difference V across R. (i) Use Fig. 7.2 to determine 1. the maximum power dissipation in R, 2. the potential difference across R when the maximum power is dissipated. (ii) Hence calculate the resistance of R when the maximum power is dissipated. (iii) Use your answers in (i) and (ii) to determine the internal resistance r of the battery. (c) By reference to Fig. 7.2, it can be seen that there are two values of potential difference V for which the power dissipation is 1.05W. State, with a reason, which value of V will result in less power being dissipated in the internal resistance. 6. The heating element of an electric kettle operates at 230 V and has a power rating of 2.52 kw.a) Calculate the working resistance R of the heating element.b) A digital multimeter is used to measure the resistance of the heating element when it is cold. The value of the resistance measured by this method is considerably lower than the working resistance.suggest a reason for this. 7. A simple cell may be constructed by inserting into a fresh lemon two electrodes made from different metals. The juice of the lemon acts as an electrolyte (conducting liquid). Positive and negative ions within the lemon move towards the metal electrodes. Fig.below shows such a lemon-cell. It has an e.m.f. of 1.32 V and can provide enough electrical energy to activate a digital lock for many days.(a) On Fig. 2.1, indicate with an arrow the direction in which negative charge moves within the lemon.(b) The lemon-cell is capable of providing a steady current of 1.2 ma for eight days ( s). Calculate (i) The charge passing through the clock during eight days (ii) the power delivered by the lemon-cell. 11

8. Two heating coils X and Y, of resistance R X and R Y respectively, deliver the same power when 12 V is applied across X and 6 V is applied across Y.What is the ratio R X /R Y? 9.

12 8. Two heating coils X and Y, of resistance R X and R Y respectively, deliver the same power when 12 V is applied across X and 6 V is applied across Y.What is the ratio R X /R Y? 9. Two wires P and Q have resistances R P and R Q respectively. Wire P is twice as long as wire Q and has twice the diameter of wire Q. The wires are made of the same material. What is the ratio R p /R Q? 10. A cylindrical piece of a soft, electrically-conducting material has resistance R. It is rolled out so that its length is doubled but its volume stays constant.what is its new resistance? 11. An electric railway locomotive has a maximum mechanical output power of 4.0 MW. Electrical power is delivered at 25 kv from overhead wires. The overall efficiency of the locomotive in converting electrical power to mechanical power is 80 percent.what is the current from the overhead wires when the locomotive is operating at its maximum power? 12. A wire carries a current of 2.0 amperes for 1.0 hour. How many electrons pass a point in the wire in this time? 13. Which electrical quantity would be the result of a calculation in which energy is divided by charge? 14. What is electric potential difference between two points in resistor? carries a current? 12

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Electricity May 02 1. The graphs show the variation with potential difference V of the current I for three circuit elements. PhysicsAndMathsTutor.com When the four lamps are connected as shown in diagram