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## Electricity

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**Electricity**N Bronks**Basic ideas…**Electric current is when electrons start to flow around a circuit. We use an _________ to measure it and it is measured in ____. Potential difference (also called _______) is how big the push on the electrons is. We use a ________ to measure it and it is measured in ______, a unit named after Volta. Resistance is anything that resists an electric current. It is measured in _____. Words: volts, amps, ohms, voltage, ammeter, voltmeter**Electrons are flowing from the negative to positive side of**the battery through the wires Note current moves from positive to negative, however electrons are actually are moving in the opposite direction!**Current**• Flow of electrons**Current and Charge**Q Charge (C) Coulomb I = Current (A) = Amp = t time (s) second Since one Ampere flows when one coulomb of charge passes a given point in a circuit each second, or Also no. of electrons x charge of one electron Charge (C) =**Example 1:**How many electrons are there in 20 Coulombs ? No. of electrons = total charge / charge of one electron No. of electrons = 20 / 1.6x10-19 No. of electrons = 1.25 x1020 electrons • Example 2: • The current in a circuit is 5A. What is the charge flowing in : • 1 second ? • 10 seconds ? 5 Coulomb 50 Coulomb**Summary Question**• a. The current in a certain wire is 0.35A. Calculate the charge passing a point in the wire i. in 10 s, ii in 10 min. • b. Calculate the average current in a wire through which a charge of 15C passes in i. 5s, ii 100s • Ans: • i. I = 0.35A t = 10s • Q = It Q = 0.35 x 10 = 3.5 C • ii. I = 0.35A t = 10 min = 600s • Q = It Q = 0.35 x 600 = 210 C • b. i. Q = 15C t = 5s • I = Q/t I = 15/5 = 3A • ii. Q = 15C t = 100s • I = Q/t I = 15/100 = 0.15A**Summary Questions page 47**2. Calculate the number of electrons passing a point in the wire in 10 minutes when the current is a. 1.0µA b. 5.0A • Ans: • I = 1.0µA t= 10min = 600s • Q = It = 1x10-6 x 600 = 6x10-4 C • No. of electrons = total charge / charge of one electron • No. of electrons = 6x10-4 / 1.6x10-19 = 3.75x1015 electrons • I = 5A t= 10min = 600s • Q = It = 5 x 600 = 3000 C • No. of electrons = total charge / charge of one electron • No. of electrons = 3000 / 1.6x10-19 = 1.88x1022 electrons**Summary Questions page 47**• 3. In an electron beam experiment, the beam current is 1.2mA. Calculate • The charge flowing along the beam each minute • The number of electrons that pass along the beam each minute • Ans: • I = 1.2x10-3 A t= 1min = 60s • Q = It = 1.2x10-3 x60 = 0.072 C • b. no. of electrons = total charge /charge of one electron • no. of electrons = 0.072/1.6x10-19 = 4.5x1017 electrons**Summary Questions page 47**• A certain type of rechargeable battery is capable of delivering a current of 0.2A for 4000s before its voltage drops and it needs to be recharged. • Calculate: • The total charge the battery can deliver before it needs to be recharged. • The maximum time it could be used for without being recharged if the current through it was i. 0.5A, ii. 0.1A • Ans: • I = 0.2A t = 4000s • Q = It = 0.2 x 4000 = 800C • i. Q = 800C I = 0.5A • t = Q/I = 800/0.5 = 1600s • ii. Q = 800C I = 0.1A • t = Q/I = 800/0.1 = 8000s**Current and charge quiz**Q I = t 1. Calculate the charge passing through a lamp in three minutes when a steady current of 0.4 A is flowing. Q = I t Q = 0.4 x 3 x 60 = 72 Coulomb**Current and charge quiz**Q I = t 2. Calculate the number of electrons flowing through a resistor when a current of 2.3 flows for 5 minutes Q = I x t Q = 2.3 x 5 x 60 = 690 Coulomb no. of electrons = 690 /1.6x10-19 = 4.31x1021**Current and charge quiz**Q I = t 3. What is the current in a circuit if 2.5x1020 electrons pass a given point every 8 seconds no. of electrons x charge of one electrons Charge (C) = 2.5x1020 x 1.6x10-19 = 40 Coulombs Charge (C) = Current = 40/8 = 5 Amps**Current and charge quiz**Q I = t 4. How long does it take for a current of 0.3A to supply a charge of 48C? t = Q/I t = 48/0.3 = 160 seconds**Current and charge quiz**Q I = t 5. How many electrons pass a point when a current of 0.4A flows for 900 seconds? Q = I x t = 0.4 x 900 = 360 Coulomb no. of electrons = total charge / charge of one electron no. of electrons = 360 / 1.6 x 10-19 = 2.25 x 1021**Current and charge quiz**6. A torch bulb passes a current of 120 mA. How many coulombs of charge flow through the lamp in 1 minute? Q = I x t = 120x10-3 x 60 = 7.2 Coulomb**Current and charge quiz**7. A car battery is rated as 36 A h. In principle this means it could pass a current of 1 A for 36 h before it runs down. How much charge passes through the battery if it is completely run down? Q = I x t = 1 x 36 x 60 x 60 = 129600 Coulomb**H/W**• 2004 HL Q4**More basic ideas…**Another battery means more current as there is a greater push on the electrons The extra resistance from the extra bulb means less current**Current in a series circuit**If the current here is 2 amps… The current here will be… The current here will be… And the current here will be… 2A 2A 2A In other words, the current in a series circuit is THE SAME at any point**Current in a parallel circuit**Here comes the current… Half of the current will go down here (assuming the bulbs are the same)… And the rest will go down here… A PARALLEL circuit is one where the current has a “choice of routes”**Current in a parallel circuit**And the current here will be… The current here will be… The current here will be… The current here will be… If the current here is 6 amps 6A 2A 2A 2A**Voltage in a series circuit**V If the voltage across the battery is 6V… …and these bulbs are all identical… V V …what will the voltage across each bulb be? Voltmeter always in parallel 2V**Voltage in a series circuit**V If the voltage across the battery is 6V… …what will the voltage across two bulbs be? V 4V**Voltage in a parallel circuit**If the voltage across the batteries is 4V… What is the voltage here? V V And here? 4V 4V**Summary**In a SERIES circuit: Current is THE SAME at any point Voltage SPLITS UP over each component In a PARALLEL circuit: Current SPLITS UP down each “strand” Voltage is THE SAME across each”strand”**An example question:**6V A3 3A A1 V1 A2 V2 V3**Advantages of parallel circuits…**• There are two main reasons why parallel circuits are used more commonly than series circuits: • Extra appliances (like bulbs) can be added without affecting the output of the others 2) When one breaks they don’t all fail**Resistance**Resistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me. That makes me so happy Georg Simon Ohm 1789-1854 V Resistance = Voltage (in V) (in ) Current (in A) I R The resistance of a component can be calculated using Ohm’s Law:**An example question:**Ammeter reads 2A A V Voltmeter reads 10V What is the resistance across this bulb? As R = volts / current = 10/2 = 5 Assuming all the bulbs are the same what is the total resistance in this circuit? Total R = 5 + 5 + 5 = 15 **More examples…**3A 3A 2A 4V 2V 1A 6V 12V What is the resistance of these bulbs?**VARIATION OF CURRENT (I) WITH P.D. (V)**A Nichrome wire + 6 V - V**Method**1. Set up the circuit as shown and set the voltage supply at 6 V d.c. 2.Adjust the by moving the slider of the potential divider to obtain different values for the voltage V and hence for the current I. 3.Obtain at least six values for V and I using the voltmeter and the ammeter. 4.Plot a graph of V against I**Variations**(a) A METALLIC CONDUCTOR With a wire (b) A FILAMENT BULB (c) COPPER SULFATE SOLUTION WITH COPPER ELECTRODES (d) SEMICONDUCTOR DIODE Done both ways with a milli-Ammeter and the a micro Ammeter**Current-voltage graphs**I I I V V V 3. Diode 1. Resistor A diode only lets current go in one direction Current increases in proportion to voltage 2. Bulb As voltage increases the bulb gets hotter and resistance increases**Factors affecting Resistance of a conductor**• Resistance depends on • Temperature • Material of conductor • Length • Cross-sectional area • Temperature • The resistance of a metallic conductor increases as the temperature increases e.g. copper • The resistance of a semiconductor/insulator decreases as the temperature increases e.g. thermistor.**VARIATION OF THE RESISTANCE OF A METALLIC CONDUCTOR WITH**TEMPERATURE 10º C 10ºC Ω Digital thermometer Wire wound on frame Water Glycerol Heat source**Method**1. Set up as shown. 2. Use the thermometer to note the temperature of the glycerol, which is also the temperature of the coil. 3. Record the resistance of the coil of wire using the ohmmeter. 4. Heat the beaker. 5. For each 10 C rise in temperature record the resistance and temperature using the ohmmeter and the thermometer. 6. Plot a graph of resistance against temperature.**Graph and Precautions**Precautions - Heat the water slowly so temperature does not rise at end of experiment -Wait until glycerol is the same temperature as water before taking a reading. R **Factors affecting Resistance of a conductor**• Length • Resistance of a uniform conductor is directly proportional to its length. • i.e. R L • Cross-sectional area • Resistance of a uniform conductor is inversely proportional to its cross-sectional area. • i.e. R 1 • A**Factors affecting Resistance of a conductor**• Material The material also affects the resistance of a conductor by a fixed amount for different materials. This is known as resistivity (). R = L = Resistivity AUnit: ohm meter m**RESISTIVITY OF THE MATERIAL OF A WIRE**Micrometer Nichrome wire Crocodile clips l Metre stick Bench clamp Stand**Method**1. Note the resistance of the leads when the crocodile clips are connected together. Could also be precaution. 2. Stretch the wire enough to remove any kinks or ‘slack’ in the wire. 3.Read the resistance of the leads plus the resistance of wire between the crocodile clips from the ohmmeter. Subtract the resistance of the leads to get R. 4.Measure the length l of the wire between the crocodile clips, with the metre stick. 5.Increase the distance between the crocodile clips. Measure the new values of R and l and tabulate the results. 6.Make a note of the zero error on the micrometer. Find the average value of the diameter d.**ρ**1. Calculate the resistivity where A = 2. Calculate the average value. Precautions Ensure wire is straight and has no kinks like .... Take the diameter of the wire at different angles**H/W**• 2004 HL Q4**Resistors in series and Parallel**I I1 I2 IT V R1 R2 R3 R1 V1 R2**Resistors in series and Parallel**I I1 I2 IT V R1 R2 R3 R1 V1 R2