Power lines. Why do birds sitting on a high-voltage power line survive?

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1 Power lines At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit high V, low I or high I, low V? (a) high V, low I (b) low V, high I (c) makes no difference Power loss is I 2 R, so want to minimize current. Why do birds sitting on a high-voltage power line survive? They are not touching high and low potential simultaneously to form a circuit that can conduct current PHY2054: Chapter 18 18

2 Resistors Current flows through a light bulb. If a wire is now connected across the bulb as shown, what happens? (a) Bulb remains at same brightness (b) Bulb dims to 1/4 its former brightness (1/2 current) (c) Bulb goes out (d) None of the above The wire shunt has almost no resistance and it is in parallel with a bulb having resistance. Therefore voltage across shunt (and bulb ) is ~ 0. Thus almost all the current follows the zero (or extremely low) resistance path. PHY2054: Chapter 18 19

3 Circuits Two light bulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, what will happen to bulb A? (a) burns more brightly than before (b) burns as brightly as before (c) burns more dimly than before (d) goes out The wire shunt effectively eliminates the second resistance, hence increasing the current in the circuit by 2x. The first bulb burns 4x brighter (I 2 R). PHY2054: Chapter 18 20

4 Series circuit Voltage Drops in Series Circuit Battery V, resistors R 1 and R 2 Current i = V / (R 1 + R 2 ) Voltage across R 1 = IR 1 = V * R 1 / (R 1 + R 2 ) Voltage across R 2 = IR 2 = V * R 2 / (R 1 + R 2 ) So voltage drops in proportion to resistance! Let V = 24, R 1 = 8Ω, R 2 = 4Ω V 1 = 24 * 8 / 12 = 16 V 1 = 24 * 4 / 12 = 8 Useful when considering how resistors divide up voltage PHY2054: Chapter 18 21

5 Circuits Consider the network of resistors shown below. When the switch S is closed, then: What happens to the voltage across R 1, R 2, R 3, R 4? What happens to the current through R 1, R 2, R 3, R 4? What happens to the total power output of the battery? Let R 1 = R 2 = R 3 = R 4 = 90 Ω and V = 45 V. Find the current through each resistor before and after closing the switch. Before I 1 = 45/135 = 1/3 I 2 = 0 I 3 = I 4 = 15/90=1/6 After I 1 = 45/120 = 3/8 I 2 = I 3 = I 4 = 1/8 PHY2054: Chapter 18 22

6 Circuits The light bulbs in the circuits below are identical. Which configuration produces more light? (a) circuit I (b) circuit II (c) both the same Circuit II has ½ current of each branch of circuit I, so each bulb is ¼ as bright. The total light in circuit I is thus 4x that of circuit II. PHY2054: Chapter 18 23

7 Circuits The three light bulbs in the circuit are identical. The current flowing through bulb B, compared to the current flowing through bulb A, is a) 4 times as much b) twice as much c) the same d) half as much e) 1/4 as much Branch of circuit A has ½ resistance, thus it has 2x current. PHY2054: Chapter 18 24

8 Circuits The three light bulbs in the circuit are identical. What is the brightness of bulb B compared to bulb A? a) 4 times as much b) twice as much c) the same d) half as much e) 1/4 as much Use P = I 2 R. Thus 2x current in A means it is 4x brighter. PHY2054: Chapter 18 25

9 More Complicated Circuits Parallel and series rules are not enough! Use Kirchoff s rules PHY2054: Chapter 18 26

10 Problem Solving Using Kirchhoff s Rules Label the current in each branch of the circuit Choice of direction is arbitrary (signs work out in the end) Apply junction rule at each junction: I = I 1 + I 2 I Apply loop rule to each loop (follow in one direction) Resistors: if loop direction in current direction, voltage drop Batteries: if loop direction in normal direction, voltage gain Sum of all voltages = 0 around loop I +V - IR Solve equations simultaneously You need as many equations as you have unknowns I 1 I 2 See example next slide PHY2054: Chapter 18 27

11 Example 1: Applying Kirchhoff s Rules Determine magnitudes and directions of all currents Take two loops, 1 and 2, as shown Choose currents & directions!: I 1, I 2, I 3 Use I 1 = I 2 + I 3 I 3 9 V 15 Ω 22 Ω 2 I I = 0 22I I = V I 1 1 I 1 I I 2 3 = 15/ 22 = 0.68 = 6/15 = 0.40 I = I + I = PHY2054: Chapter 18 28

12 Example 2: Applying Kirchhoff s Rules Pick currents as shown I 3 = I 1 + I I I = I 2I = I I 1 2 = 2 = 3 I = I + I = PHY2054: Chapter 18 29

13 Circuits Which of the equations is valid for the circuit loop below? a) 2 I 1 2I 2 = 0 b) 2 2I 1 2I 2 4I 3 = 0 c) 2 I 1 4 2I 2 = 0 d) I 3 2I 2 4I 3 = 0 e) 2 2I 1 2I 2 4I 3 = 0 PHY2054: Chapter 18 30

14 Circuit Problem (1) The light bulbs in the circuit are identical. What happens when the switch is closed? a) both bulbs go out b) the intensity of both bulbs increases c) the intensity of both bulbs decreases d) nothing changes Before: Potential at (a) is 12V, but so is potential at (b) because equal resistance divides 24V in half. When the switch is closed, nothing will change since (a) and (b) are still at same potential. (a) (b) PHY2054: Chapter 18 31

15 Circuit Problem (2) The light bulbs in the circuit shown below are identical. When the switch is closed, what happens to the intensity of the light bulbs? a) bulb A increases b) bulb A decreases c) bulb B increases d) bulb B decreases e) nothing changes (a) (b) Before: Potential at (a) is 12V, but so is potential at (b) because equal resistance divides 24V in half. When the switch is closed, nothing will change since (a) and (b) are still at same potential. PHY2054: Chapter 18 32

16 Circuit Problem (3) The bulbs A and B have the same R. What happens when the switch is closed? a) nothing happens b) A gets brighter, B dimmer c) B gets brighter, A dimmer d) both go out (a) (b) Before: Bulb A and bulb B both have 18V across them. 24V After: Bulb A has 12V across it and bulb B has 24V across it (these potentials are forced by the batteries). PHY2054: Chapter 18 33