PHYSICS 151 Notes for Online Lecture #38

Transcription

1 PHYSCS 5 Notes for Online Lecture #38 Power Power is defined as the energy transformed/time. P Energy time When a charge, q, sses across a otential difference,, it acquires an energy q. f it takes a time t to ss across the otential difference, the ower is q F q P t H G t K J Where we ve used that q/t is the current. The ower is thus given by. The unit of Power, the Watt, is equivalent to an A- We can also write the ower in terms of the resistance, by using Ohm s law: P P ( ) P F H G K J These two latter forms aly to things that have resistors and the first one alies in general. Ex. : What is the resistance of a 00-W light bulb oerating at 0? P P ( 0) 44Ω 00W Ex. : What is the current ssing through a 00 W hair dryer oerating at 0? Here, we use the P form: P P 00W 0A 0 Electric Circuits Page

2 Ex. 3: f electricity is $0.05 er kw-hr, how much does it cost to run a 00 W light bulb for hour? First, what is a kwh it s not a S unit, so it s not obvious whether it is an energy or ower or what. Using kw 000 W and h 3600 s, kwh (000W)(3600s)3.6x0 6 J A kw-hr is thus a unit of energy. From Power Energy/time, E Pt We now have to convert this to kwh E F H G K J b g. J s J 36x0 5 J s 5 E 36. x0 J F H G E 0. kwh kwh x0 J so the total energy used for the light bulb above will be 0. kwh. At $0.05 er kwh, it costs about $0.005 to run the light bulb for an hour. Current roduces heat. A current ssing through an object with a resistance will roduce heat. Light bulb filaments are made of tungsten. As you increase the amount of current ssing through the bulb, the filament glows brighter and brighter. Although some of the ower goes to roducing light, most of the ower goes to roducing heat. This is the rincile uon which your hair dryer works. The ower dissited within the tungsten filament is P f the resistance is very large, the ower is large and the wire will heat u and could melt or start a fire. KJ Electric Circuits Page

3 Kirchoff s ules and Simle esistive Circuits f you buy chea Christmas tree lights, you will find that when one lam goes out, all of the lights go out. This has to do with how the current is delivered to the different lights. n order to understand this henomena, we need to understand how current behaves when confronted with multile resistors. There are two fundamental ideas that have to be followed. ) energy must be conserved in the circuit ) charge must be conserved These two ideas are exressed in Kirchoff s ules. The first rule conservation of energy is exressed as: The algebraic sum of the otential differences around a closed conducting loo must be zero. This is sometimes known as Kirchoff s voltage rule or the loo rule. The second rule conservation of charge says The net current into any junction must be zero n other words, all of the current (electrons) that comes into a junction must leave the junction. This is sometimes called Kirchoff s current rule, or the junction rule. We re going to look at several situations to discover how combinations of resistance and batteries can be exlained. Batteries in Series n some situations, we might want to ut multile batteries together (say, for instance, in a flashlight or radio.). f take a single flashlight battery, the otential difference across its terminals is.5. Now imagine that hook two batteries together, with the ositive terminal of one battery hooked to the negative of the other. What is the voltage then? n this situation, the voltages of the individual batteries add, so that the total voltage is The otential at oint A is 0, 3.0 at oint B is.5 and at.5 oint C is 3.0. difference A B C Note that if hook the batteries u in the same way, but ut the ositive terminals together, don t get any otential Note that batteries laced in rallel do not increase the voltage, however they allow current to be roduced for a longer time. Electric Circuits Page 3

4 esistors in Series B A C D E Consider a circuit consisting of two resistors, and and a battery with a voltage. The current through resistors in series is the same. f you think about the current traveling from the ositive terminal of the battery, there is only one th for the current to take. This means that the current through both of the resistors is the same. We ll call it. The otential dro across the first resistor is given by Ohm s Law: and the otential across the second resistor is Kirchoff s voltage rule states that the algebraic sum of all of the otential differences must be zero around a closed loo. f you start from the negative terminal of the battery, the otential there is zero (ground). Going across the ositive terminal, you get a ositive otential difference of. The battery makes a ositive contribution to the overall otential difference. When you move across the resistor, there is a otential dro, which means that the otential on the right side of the resistor is lower than the otential on the left side of the resistor. Going around the circle, we can make the following table: Potential A 0 B C f we write Kirchoff s voltage rule, we get: We can solve this equation for D E 0 0 ( + ) s Electric Circuits Page 4

5 where s is the resistance of the two resistors in series. Two resistors in series can thus be considered as a single resistance with a magnitude equal to the sum of the two. This works for as many resistors as you have, rovided they are all hooked u in series. s esistors in Parallel n some cases, we want to hook u electrical resistances in rallel, as shown to the left. Let the battery again have voltage. C D F B A E G H This time, if you think about the current, you see that when the current reaches oint C, it has an otion. t can either go toward oint D or toward oint F. You can use the water through a ie analogy again. The larger the resistance, the smaller diameter ie. More water will go through the ie having less-resistance (the wider one). Similarly, more current will go through the less-resistive resistor. When the current reaches oint C, it will slit. Kirchoff s second rule tells us that the sum of the current going into a junction lus the current leaving the junction must equal zero. f we assign the different currents names as we aroach junction C, as shown below, Kirchoff s second rule requires that + There will be a otential dro across each resistor. The otential dro across will be given by where Similarly, D B The voltage across each branch of a rallel circuit is the same. Comre the otential differences across oints C to H. egardless of which th you choose, oint C starts C out with some otential (in this case, ) and must have some other otential at oint H. This means that the otential dro across is the same as the otential dro across. F G This means that the equations reresenting the otential dro across the two resistors are: and we can solve each of these for the current: A E H Electric Circuits Page 5

6 Plug these values into Kirchoff s current rule: + + F HG + Solving for, F + HG KJ KJ where is the equivalent resistance for the two resistors in rallel. n the general case, the equivalent resistance of resistors connected in rallel is: When resistors are in series, the effective resistance is greater than any of the individual resistors. When resistors are in rallel, the effective resistance is less than any of the individual resistances Electric Circuits Page 6

7 Ex. 4: Analyze the following circuit if 8Ω, 6Ω, 3 3 Ω, and 0. First, find the effective resistance of the rallel circuit + + 6Ω 3Ω 3 6 Ω 6 Ω 3 Ω 3 3 Now what you have is in series with Add these two resistors in series: TOTAL + 8Ω + Ω 0Ω So the current leaving the battery is A and this must also be the current through. However, when the current gets to the junction where the two ths lead to and 3 it must divide. Note that when the current sses through there is a voltage dro of (A)(8Ω) 6. So the voltage dro across both and 3 must be 4. So (4) (6 Ω) 0.67 A (4) (3 Ω) 3.33 A Note that + 3 which we know must be true. Electric Circuits Page 7

8 Ex. 5: For the circuit shown to the right, find: a) the current ssing through each resistor b) the voltage dro across each resistor B C H 3 Take: /3 Ω Ω 3 3 Ω A D E F G J 4 5 K L 4 4 Ω 5 Ω Solution: First, find the equivalent resistances for the rallel combination involving and Ω Ω 3 + Ω Ω 4 Ω Ω 4 Ω Now, find the equivalent resistance for the rallel combination involving 3, 4 and 5 : Electric Circuits Page 8

9 From this, we can find the current that flows through the circuit. + + b b 3Ω 4 Ω Ω b Ω Ω Ω 8 b Ω b Ω 8 The circuit can now be relaced by its equivalent the total resistance is thus the sum of the two resistances in series T + b T Ω+ Ω 8 4 T Ω+ Ω T Ω 8 Ω 6A Ω Note that this is the current going into each of the combinations of resistors in rallel: i.e. 6 A flows from A to F and from G to L. The otential dro over each combination can now be calculated: AF b A F G L a fa f AF 6A Ω 3 We can calculate GL one of two ways. First, we know that the sum of the voltage dros across the circuit must algebraically add to zero, so if the total voltage is and the otential dro from A to F is 3, the dro from G to L must be 9. We can also calculate it directly: T A B C F D E Electric Circuits Page 9

10 GL b a fb GL 6A 8 Ω 9 But these calculations don t tell us how much current and voltage are going through each resistor. To do that, we have to look more carefully at the rallel combinations. We know from our above calculation that AF 3. We know that the same voltage dro will occur over each branch of the rallel circuit, so BC DE AF 3 The currents across and are then given by Ohm s Law: BC A 3 Ω DE 3 5A Ω. We can check this. We know that 6 A of current sses from oint A to oint F, so the sum of the currents through and must be equal to 6 A. + 6 A 45. A+ 5. A 6A The same tye of calculation can be erformed for the combination that makes u b. The same voltage dro will occur over each branch of the rallel circuit, so H GL JK 9 3 The currents across 3, 4 and 5 are then given by Ohm s H Law: 4 3 H 9 3A 3Ω 3 GL 9 5. A 4Ω 4 JK A 5 5 Ω We can check this. We know that 6 A of current sses from oint A to oint F, so the sum of the currents through and must be equal to 6 A. + 6 A 3A+ 5. A A 6A g G J 4 5 K L n summary esistor Current (A) oltage () Electric Circuits Page 0

11 /3 Ω Ω Ω Ω Ω Electric Circuits Page