A –Level Physics: Electrical Quantities: Internal Resistance

Objectives: Additional skills gained: Practical Planning 45. Know the definition of electromotive force (e.m.f.) and understand what is meant by internal resistance and know how to distinguish between e.m.f. and terminal potential difference 47. Understand how changes of resistance with temperature may be modelled in terms of lattice vibrations and number of conduction electrons and understand how to apply this model to metallic conductors and negative temperature coefficient thermistors 48. Understand how changes of resistance with illumination may be modelled in terms of the number of conduction electrons and understand how to apply this model to LDRs. Additional skills gained: Practical Planning Integrating GCSE content

Starter: Riddle (answer revealed next lesson) You have one 16-quart pitcher full of water, one empty nine-quart pitcher, and one empty seven-quart pitcher… Using nothing but these three pitchers, can you dividethe water evenly so that the 16-and nine-quart pitchers are each holding exactly eight quarts of water? 10mins 16 9 7 21978

Starter: Riddle (answer revealed next lesson) Solution : Pour from... To... Quarts... 1) 16-quart pitcher 9-quart pitcher ( 7/9/0) 2) 9-quart pitcher 7-quart pitcher ( 7/2/7) 3) 7-quart pitcher 16-quart pitcher (14/2/0) 4) 9-quart pitcher 7-quart pitcher (14/0/2) 5) 16-quart pitcher 9-quart pitcher ( 5/9/2) 6) 9-quart pitcher 7-quart pitcher ( 5/4/7) 7) 7-quart pitcher 16-quart pitcher (12/4/0) 8) 9-quart pitcher 7-quart pitcher (12/0/4) 9) 16-quart pitcher 9-quart pitcher ( 3/9/4) 10) 9-quart pitcher 7-quart pitcher ( 3/6/7) 11) 7-quart pitcher 16-quart pitcher (10/6/0) 12) 9-quart pitcher 7-quart pitcher (10/0/6) 13) 16-quart pitcher 9-quart pitcher ( 1/9/6) 14) 9-quart pitcher 7-quart pitcher ( 1/8/7) 15) 7-quart pitcher 16-quart pitcher ( 8/8/0) 21978

This resistor would have a very low value. Internal Resistance Resistance quantifies how strongly a component opposes the flow of charge carriers (current) Why is it that when a voltmeter is placed around a 1.5V battery, it doesn’t read ‘1.5v’? When a current is flowing through a circuit, the electrons still have to travel through the cell. A small portion of the voltage/energy will be lost here because the electrons have to overcome the internal resistance of the cell! Conduction is the movement of charged particles or propagation of energy through a material Because the cell has this internal resistance, the convention is to show the cell adjacent to a resistor. This resistor would have a very low value. Delta

P= I x V (so we can substitute ‘V’ for I x R) Power/I.Resistance equations What does V= in Ohm’s law? So, if Power=Current x Voltage, what is the formula relationship between Power and resistance? V= I x R P= I x V (so we can substitute ‘V’ for I x R) P= I x I x R P= I2 x R As electrons are resisted through a cell, they will dissipate power (work done in a set time). This is why each coulomb, will not get the full number of joules (voltage). So, the EMF is actually what is left when we subtract the voltage used in internal resistance. So as V=IR, what is the equation for the EMF of this circuit? Delta

Power/I.Resistance equations So the total energy produced by the cell (ε). Is the voltage used against internal resistance added to the voltage supplied to the rest of the circuit. So what’s the equation? ε =IR + Irinternal So, rearranging: IR = ε - Irinternal V= ε – Irinternal So this shows that the voltage produced at the terminals of the cell (i.e. outside of the cell) is effectively the EMF – the voltage overcoming internal resistance Delta

Typical Exam Q #1 Q1. A battery is connected to a 10ohm resistor as shown. The e.m.f. (electromotive force) of the battery is 12V. (a) (i) Explain what is meant by the e.m.f. of a battery. (ii) When the switch is open the voltmeter reads 12.0V and when it is closed it reads 11.5 V. Explain why the readings are different. (b) Calculate the internal resistance of the battery. (a) (i) Explain what is meant by the e.m.f. of a battery. The EMF is the energy provided by a battery  per unit charge  (OR when no current flows it is the potential difference across the battery ) (ii) When the switch is open the voltmeter reads 12.0V and when it is closed it reads 11.5 V. Explain why the readings are different. A potential difference is developed across the internal resistance  (3) (b) Calculate the internal resistance of the battery. I =11.5/10 = 1.15A  'lost volts' = Ir 12 – 11.5 = 1.15 × r  r = 0.43   OR EMF/I = R + r 12/1.15 =10 + r  10.43 - 10 = r = 0.43  

Typical Exam Q #2 Q2. A very high resistance voltmeter reads 20V when connected across the terminals of a d.c. power supply. The high resistance meter is disconnected and a second voltmeter of resistance 1.0k is then connected across the supply. The second meter gives a reading of 16V. (i) State the e.m.f. of the power supply. (ii) Calculate the current which flows through the second meter. (iii) Calculate the internal resistance of the power supply. (iv) Show that the current is equal to 0.080A when the supply is short circuited. A very high resistance voltmeter reads 20V when connected across the terminals of a d.c. power supply. The high resistance meter is disconnected and a second voltmeter of resistance 1.0k is then connected across the supply. The second meter gives a reading of 16V. (i) State the e.m.f. of the power supply. 20V  (ii) Calculate the current which flows through the second meter. V = IR I = V/R = 16/1000 = 1.6 × 10–2 A  (iii) Calculate the internal resistance of the power supply. 'lost volts' = Ir 20 - 16 = 1.6 × 10–2 × r  r = 250   OR EMF/I = R + r 20/1.6 × 10–2 =1000 + r  r = 250   (iv) Show that the current is equal to 0.080A when the supply is short circuited.Short circuited - no external resistance EMF = Ir I = 20/250 = 8.0 × 10–2 A

Typical Exam Q #3 Q3 (a) A steady current of 0.25 A passes through a torch bulb for 6 minutes. Calculate the charge which flows through the bulb in this time. (b) The torch bulb is now connected to a battery of negligible internal resistance. The battery supplies a steady current of 0.25 A for 20 hours. In this time the energy transferred in the bulb is 9.0 × 104 J. Calculate (i) the potential difference across the bulb, (ii) the power of the bulb A very high resistance voltmeter reads 20V when connected across the terminals of a d.c. power supply. The high resistance meter is disconnected and a second voltmeter of resistance 1.0k is then connected across the supply. The second meter gives a reading of 16V. (i) State the e.m.f. of the power supply. 20V  (ii) Calculate the current which flows through the second meter. V = IR I = V/R = 16/1000 = 1.6 × 10–2 A  (iii) Calculate the internal resistance of the power supply. 'lost volts' = Ir 20 - 16 = 1.6 × 10–2 × r  r = 250   OR EMF/I = R + r 20/1.6 × 10–2 =1000 + r  r = 250   (iv) Show that the current is equal to 0.080A when the supply is short circuited.Short circuited - no external resistance EMF = Ir I = 20/250 = 8.0 × 10–2 A

Typical Exam Q #4 Q4 (a) A student is given three resistors of resistance 3.0 , 4.0  and 6.0 ohms  respectively. (i) Draw the arrangement, using all three resistors, which will give the largest resistance. (ii) Calculate the resistance of the arrangement you have drawn. (iii) Draw the arrangement, using all three resistors, which will give the smallest resistance. (iv) Calculate the resistance of the arrangement you have drawn. (b) The three resistors are now connected to a battery of EMF 12 V and negligible internal resistance, as shown in the diagram above (ii) Calculate the voltage across the 6.0  resistor a) A student is given three resistors of resistance 3.0 , 4.0  and 6.0  respectively. (i) Draw the arrangement, using all three resistors, which will give the largest resistance. The largest resistance will be the three in series.  (ii) Calculate the resistance of the arrangement you have drawn. R = 3.0 + 4.0 + 6.0 = 13   (iii) Draw the arrangement, using all three resistors, which will give the smallest resistance. The smallest resistance will be the three in parallel.  (iv) Calculate the resistance of the arrangement you have drawn. 1/R = 1/3.0 + 1/4.0 + 1/6.0 = (4.0 + 3.0 + 2.0)/12 = 9.0/12 = 0.75  R = 1/0.75 = 1.3   (5) (b) The three resistors are now connected to a battery of EMF 12 V and negligible internal resistance, as shown in the diagram below. (i) Calculate the total resistance in the circuit. First we have to replace the parallel arrangement with a single resistor. 1/R = 1/3.0 + 1/6.0 = 1/2.0 Rparallel = 2.0   We now have a series arrangement RTotal = 2.0 + 4.0 = 6.0   (ii) Calculate the voltage across the 6.0  resistor. The 6.0  resistor is part of the parallel arrangement that has a resistance of 2.0 . It will therefore get that share of the voltage. The 12V from the battery is shared out across 6.0 . Each ohm gets 2.0 V - the parallel arrangement will therefore get 4.0 V and the voltage across the 6.0  resistor will be 4.0 V 

Typical Exam Q #5 Q5. In the circuit shown below, the battery, of emf 6.0V, has negligible internal resistance. (a) Calculate the current through the ammeter when the switch S is (i) open (ii) Closed (b) The switch S is now replaced with a voltmeter of infinite resistance. Determine the reading on the voltmeter. 8. In the circuit shown below, the battery, of emf 6.0V, has negligible internal resistance. (a) Calculate the current through the ammeter when the switch S is (i) open, R = 20 + 60= 80  V = IR so I =V/R I = 6.0/80 = 0 075 A  (ii) closed When the switch is closed the 60  resistor is 'shorted out'. It is in effect a resistor in parallel with a piece of connecting wire. The 'short' wire and resistance parallel combination have less resistance than the wire itself - therefore can be ignored! It is as if there is just a piece of connecting wire at that point in the circuit. R = 20   I =V/R = 6.0/20 = 0.3 A  (3 MAX) (b) The switch S is now replaced with a voltmeter of infinite resistance. Determine the reading on the voltmeter. The voltmeter has such a high resistance that the parallel arrangement it makes with the resistor has a value of the resistor alone (remember that the reciprocal of infinity is zero, so using the 'resistors in parallel' equation, you get that result. We now have the same circuit as in (a)(i), but with the addition of the voltmeter. The voltmeter will have the same voltage across it as the 60 ,  V = IR = 0.075 × 60 = 4.5 V  OR the voltmeter will read 60/(20 + 60) = 3/4 of the total voltage  3/4 x 6 = 4.5 V 

Grab a laptop and go to the following URL: Practice Questions Grab a laptop and go to the following URL: http://www.cyberphysics.co.uk/Q&A/KS5/electricity/circuits/circuits.html For each question: Answer the question with DETAILED working out (step by step how you’re working it out) not just the numbers Once completed, then check the solution (bottom right of each question) and mark/correct 40m Extension: http://www.cyberphysics.co.uk/Q&A/KS5/electricity/resistivity/index_resistivity.html

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