Unit 2. Current, Voltage and Resistance

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1 Strand G. Electricity Unit 2. Current, Voltage and Resistance Contents Page Current 2 Potential Difference, Electromotive Force and Power 5 Resistance and Ohm s Law 9

2 G.2.1. Current In a metallic conductor such as copper, the charge carriers are electrons. In the electrostatic case we considered these charge carriers to be stationary, although this not actually the case. The electrons in the conductor are actually moving at speeds of up to 10 6 m/s, orbiting and being swapped between atoms. However, since their motion is completely random in terms of direction, they do not go anywhere overall. When an electric field is applied across the conductor the electrons experience a force F = qe along the electric field lines. This results in an overall acceleration in a given direction. The electrons in the conductor still move randomly at high speeds, but with an overall velocity in the direction opposite to the E field as shown by Figure G This overall velocity due to the applied E field is the drift velocity vd, and is very slow (a few millimetres a second). So if electrons only travel a few millimetres a second in an electrical wire, why does a light bulb light up the instant that the switch (which may be many metres from the light) is turned on? The reason is that the E field is set up in the wire at a speed close to the speed of light, and therefore electrons start to move all along the wire at approximately the same instant. Path followed by an electron in a conductor with no applied E field P1 P2 vd t P3 Path followed by an electron in a conductor with an applied E field, resulting in a net displacement vd t Figure G When an electrical circuit is connected, for example a set of wires passing through a lamp which are connected to a battery, the battery supplies an E field in the wire, it does NOT supply the electrons. Batteries are NOT tiny electron making machines! The E field supplied by the battery causes the charges to start flowing around the circuit. The flow of charge is current, I. Current I = the flow of charge Q The size of the current is a measure of the rate of flow of charge. CCCCCCCCCCCCCC II = CChaaaaaaaa FFFFFFFF TTTTTTTT TTTTTTTTTT or in symbols II = QQ tt 2

3 The unit of charge is the Ampere (A). 1 Ampere is equal to the flow of 1 Coulomb of charge per second. Since there is C of charge carried by every electron, 1 Ampere is also 1/ = electrons flowing past a point in the wire per second. In this way, A current of 1A flowing for 10s = 1 10= 10C of charge. A current of 5A flowing for 10s = 10 10= 100C of charge. Worked Example A kettle uses a current of 13A. It takes 3 minutes to boil. How much charge in Coulombs, and how many electrons, flowed through the kettle in this time? Answer The current of 13A flowed for 180s. Since the amount of charge that flowed is II = QQ tt QQ = II tt = 13AA 180ss = 2340CC Each electron carries a magnitude of C, therefore 2340 = So in the time it took for the kettle to boil, electrons flowed through it. In different substances, the sign of the charge carriers may be either positive or negative. For example, in a salt solution, or an ionized gas, charge carriers can be both positive ions, and negative electrons. In a metal however, charge carriers are always negative (electrons). I + - Lamp As we learnt in Unit 1 of this Strand, positive charge carriers flow in the direction of the applied E field whilst negative charge carriers flow in the direction opposite to the E field. When current rules were established (in around 1820) it was thought that the charge carriers in a metal were positive. Thus current direction, known as conventional current and usually shown by the symbol I and an arrow, as in Figure G.2.1.2, points vd - vd - I vd vd Figure G E 3

4 in the direction of positive charge carrier flow. As such the electron flow in a metal is always in the opposite direction to the current. Note from Figure G that conventional current always flows from the +ve terminal to the ve terminal of a battery, in the same direction as the E field. The electrons therefore flow from the ve to the +ve terminal. Exercise G A torch uses a 3A bulb. If the torch is used for 30 seconds, how many Coulombs of charge flows through the bulb? How many electrons? 2. Dave has purchased a lava lamp from a retro high street store. He plugs it in and waits in opened mouthed awe for four minutes, for a large cigar shaped blob of red wax to start floating to the top of the lamp and sink back down again. If 2880C of charge flowed through the lamp during this time, what current does the lamp draw? 3. Copy the following circuit diagram for the lava lamp. Mark the conventional current, and the direction of electron flow. Power Supply + - Lamp 4. Laurens hair dryer uses a current of 8A. In the time it took her to dry her hair, 3840C of charge passed through the hair dryer. How many electrons flowed through the hair dryer? How long did it take Lauren to dry her hair? Challenge Question 5. A laptop battery supplies a current of 3A for 4 hours before it needs recharging. Calculate; (a) The total charge the battery can deliver on a single charge (b) The maximum time the battery could be used if the current supplied was altered in the energy saving preferences to (i) 2A (ii) 0.8A 4

5 G.2.2 Potential Difference, Electromotive Force and Power. Every electrical circuit is a loop containing charges. The amount of charge in the loop is equal to the number of free charges in the loop. Charges are not used up by components in the circuit, and are not created by the power source. Therefore, the amount of charge in a particular circuit always remains the same. This is the Conservation of Charge. Charge cannot be created or destroyed. Therefore, the net charge of a closed system or circuit loop is a constant. In every circuit, the charges must be pushed around the circuit, in order to deliver energy to a component in the loop. Therefore, there must always be a device somewhere in the loop that acts like the water pump in a water fountain, supplying energy. This device is the battery or power supply. The push that the battery or supply provides is the electromotive force, abbreviated to emf (ε). It s a little unfortunate that the emf isn t actually a force but energy per unit charge, with the unit of JC -1 or Volt (V). The emf of a source is defined as the electrical energy produced per unit charge passing through the source For a battery driven circuit, as the electrons are pushed around the circuit they enter the battery and are supplied with a fixed amount of electrical potential energy. This energy supplied per unit charge is the emf. The electrons then leave the battery, continuing around the circuit loop until they encounter a component, such as a lamp, and deliver their energy. The electrons then continue on to the battery where their electric potential is replenished and the process repeated. The energy is delivered to the component by the electron doing work in order to pass through the component. The work done W is equal to the loss of potential energy of the electron. This potential difference (which is abbreviated to p.d, or more often called the voltage (V)) is therefore equal to the work done on, or energy transferred to, the component. Potential difference (P.d) or voltage (V) is the work done (or energy transferred) per unit charge. If work W is done when a charge Q flows through a component, the p.d (V) across the component is; VV = WW QQ 5

6 Just like emf, the unit of potential difference is the Joule per Coulomb, or Volt (V), since they are both a measure of electrical energy per unit charge. So what is the difference between the emf and the p.d? The emf (ε) is the energy per unit charge supplied to the circuit. The potential difference (p.d) or voltage is the energy used by, or the work done on components in the circuit. Remember: Emf s are energy given to the circuit. p.d s are energy taken from the circuit. Worked Example: (a) If 120J of energy is transferred to a lamp when 15C of charge passes through it, what is the potential difference across the lamp? (b) If the p.d. across a lamp is 3V, how much energy would 6C of charge transfer to the lamp? Answer (a). The energy transferred is equal to the work done, therefore W = 120J. The charge Q passing through the lamp is 15C. Since V = W/Q, VV = 120JJ 15CC = 8VV (b) The p.d (V) across the lamp is now 3V and Q = 6C. Therefore WW = QQQQ = 3VV 6CC = 18JJ When an electrical component is connected in a circuit through which current flows, there is always a potential difference across its terminals, since the electrons have higher electric potential before they enter the device, and lower when they exit, the difference in the potential is the work done by the charge on the component. If the current flowing is I, in a time t The charge that has flowed through the component is Q=I t The work done by the charge carriers is W = QV, and subbing I t for Q, W = IV t Since the work done is equal to energy transferred, and power = energy transferred / time, 6

7 Thus PP = IIIIΔtt Δtt = IIII Electrical Power = Current Voltage or in symbols PP = IIII The unit of power is the Watt (W). As electrical power is the product of current and voltage, 1W=1 Js-1 = 1VA. In addition, 1 volt = 1W per Ampere, so when the p.d is 6V across a component, 6W of work is done on the component per ampere of current. Worked Example Much to the annoyance of the science teacher, the students studying electrical circuits choose to connect a 6-volt buzzer to their circuit rather than a lamp. If the buzzer is rated at 10W, and the circuit is connected to two 3V batteries, calculate (a) The current through the buzzer. (b) The electrical energy transferred to the buzzer in an ear splitting 3 minutes. Answer Power of the buzzer P = 10W. The pd or voltage of the buzzer is 6V. Since P = IV, II = PP VV = 10WW 6VV Power P = E/t, and t = 180s. Therefore EE tt = 1.7AA = IIII EE = IIIIII = 1.7AA 6VV 180ss = 1836JJ 7

8 Exercise G Hilary has set up a train set for her son Freddie. The metal train wheels complete the circuit to the trains electric motor by bridging the metal tracks, and at full power 30J of energy is transferred every second when 5 Coulombs of charge passes through it. Calculate the voltage across the train tracks. 2. A 230V vacuum cleaner has a power rating of 800W. Calculate; (a) The energy transfer in the vacuum cleaner in 3 minutes (b) The current drawn by the vacuum cleaner 3. If an 8W lamp is connected to a circuit supplying a current of 2A, calculate the voltage across the terminals of the lamp. 4. If an electric heater draws 5A from the mains supply (230V) when operating at full power, calculate the power rating of the device to the nearest Watt. Challenge Question 5. A stereo system draws a current of 3A when connected to a mains supply of 230V. If the stereo uses a total of Joules of energy while playing the 1972 single Starman by David Bowie, how long is the track in minutes and seconds? 8

9 G.2.3 Resistance and Ohm s Law As electrons flow through a wire or a component, they do not travel unimpeded. Instead they continually bump into other charge carriers and collide with the ion cores (the materials atoms that have given up electrons to be charge carriers) which are vibrating about fixed points throughout the material. The resistance that the electrons feel from these collisions reduces their rate of progress (current) for a given voltage. We conclude that the greater resistance for a fixed voltage, the smaller the current. RRRRRRRRRRRRRRRRRRRR = PPPPPPPPPPPPPPPPPP DDDDDDDDDDDDDDDDDDDD CCCCCCCCCCCCCC or in symbols RR = VV II The unit of resistance is the Ohm (Ω), which is equivalent to one volt per ampere. Worked Example A student in a science class connects a high resistance wire to a variable power supply. He sets the power supply to 8V, and measures the current in the wire to be 2.4mA. Calculate the resistance in the wire. The student then increases the voltage to 12V. What is the current in the wire at this voltage? Answer For the 8V setting, V = 8V I = 2.4mA = A. Therefore RR = VV II = 8VV = Ω = 3.3kkΩ AA The student then increases the voltage to 12V for the same wire. Since R=V/I II = VV RR = 12VV Ω = AA = 3.6mmmm When measuring the resistance in a component or wire, we measure the current and voltage across the component or wire just like the student in the worked example. If voltage is varied, a plot of V vs. I may be obtained, as shown by Figure G For a wire, and for some components, the plot of V vs. I gives a straight 9

10 line through the origin, the gradient of which is the resistance. In this case the p.d across a wire is proportional to current, and the resistance is constant. If a material or component behaves in this way, it is said to be an Ohmic material since it obeys Ohm s Law; V (V) Ohm s Law states that the p.d across a metallic conductor is proportional to the current through it where V= potential difference I = current R = resistance V = IR I (A) Figure G Ohm s Law is equivalent to saying that voltage is directly proportional to current, the constant of proportionality being the unchanging resistance. For a resistance to be a constant however, the material or component must be held at constant temperature. NOTE: V=IR may only be used for Ohmic materials since it assumes a linear relationship between current and voltage. The resistance of a material or component is temperature dependent since, if a material is heated, the atoms or ions of the material gain energy and vibrate more vigorously with greater amplitude about their equilibrium positions. This increase in vibrational amplitude means that the flowing electrons are more likely to collide with the vibrating ions, greatly impeding their flow and increasing resistance. This in turn alters the I-V characteristics of the material. A good example of this is the standard electric light bulb. A light bulb contains a very high resistance I wire held in an inert (non-reactive) gas. Current is passed through the wire and the electrons do work in order to pass through it. The work done heats the wire until it glows white hot, producing usable light. V The I-V characteristics of a light bulb are shown in Figure (recall that for the V-I graph the gradient or rate of change of the trend line is the resistance R, therefore the gradient of an I-V graph is 1/R). As positive current begins to flow through the high Figure resistance wire of the lamp the wire behaves in accordance with Ohm s Law. However, as the current increases more energy is dissipated through charge carrier collisions, heating the wire. This causes greater vibrations of the ion cores in the wire, which leads to an increase in collisions and an increase in resistance. Since resistance is increasing, the gradient (1/R) of the I-V curve decreases, and the filament bulb 10

11 does not follow Ohm s Law. The same is true if the current is reversed (negative values of the I-V curve). Most resistors are designed to remain at a constant resistance and provide an I-V characteristic as shown by Figure (a). Some resistors however are designed to change their resistance with temperature or with exposure to light. Temperature dependent resistors are called thermistors, and resistors that increase with light exposure are light dependent resistors LDR s. These variable resistors allow changes in the environment to alter the characteristics of a circuit. For example, a light that comes on automatically at night is switched on by the change in resistance of an LDR. The I-V characteristic of a thermistor is shown in Figure 2.3.3(b). Note that the I-V characteristics of the fixed resistor, the thermistor, and an LDR follows Ohm s Law. For a conductor such as a wire as shown in Figure 2.3.4, the overall resistance depends upon the length L of the conductor, the cross sectional area, and the density of the material. The greater the length L, the greater the resistance since the further the current travels, increasing the number of collisions. Thus R is directly proportional to L. The smaller the cross sectional area the greater the resistance of the conductor for a given current, since the electrons are L confined to flow through a smaller area, increasing the rate of collisions. Thus the resistance is inversely Resistance R proportional to the cross sectional area A. The greater the density of the material the greater the resistance A since the ion cores are closer together, increasing the Figure rate of collisions. Hence R is directly proportional to material density. For a given material, RR = ρρρρ AA where ρ is a constant for that material, known as the resistivity of the material. The resistivity of a material is particular to that material, and has units of Ohm meters (Ωm). Rearranging the above equation, (a) (b) I I Figure V High Temperature Low Temperature V 11

12 ρρ = RRRR LL where ρ is the resistivity of the conductive material R is the resistance A is the cross sectional area of the conductor L is the length of the conductor Some resistivity values of example materials at room temperature are shown in Table Table Material Resistivity (Ωm) Silver Copper Aluminium Lead Silicon Pyrex glass Quartz Exercise A wire has a current of 3A flowing through it when connected to a 12V battery. Calculate the resistance of the wire. 2. A single 1kΩ resistor is placed in a circuit. Calculate the voltage across the resistor when a current of 8mA flows through it. 3. A lamp connected to a 6V battery has a current of 0.5A flowing through it. Calculate the resistance of the lamp. 4. An electrical company has a choice of using aluminium or copper to make some electrical wire for a circuit. The company want the wire to be a minimum resistance. The wire must be 50cm long, and 1mm in diameter. Use the information from Table to calculate the resistance of both the copper and the aluminium wire. How could the company reduce the resistance of the wire even further? Challenge Question 5. A light dependent resistor (LDR) is connected to an outside light as a sensor, and there is a constant potential difference of 15V across it. In the light of the day, the resistance of the LDR is 2kΩ. Calculate the current flowing through the LDR. The light requires a minimum current of 0.1A to operate. What is the maximum value of resistance the LDR can take at night when the light needs to operate within normal parameters? 12

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