PHY2054: Chapter 1818Power lines ÎAt 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 differenceÎWhy 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 currentPower loss is I2R, so wantto minimize current.PHY2054: Chapter 1819ResistorsÎCurrent 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 aboveThe wire “shunt” has almost no resistance andit 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 1820CircuitsÎTwo 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 outThe wire shunt effectively eliminates thesecond resistance, hence increasing thecurrent in the circuit by 2x. The firstbulb burns 4x brighter (I2 R).PHY2054: Chapter 1821Voltage Drops in Series CircuitÎ Series circuit Battery V, resistors R1and R2Î Current i = V / (R1+ R2) Voltage across R1= IR1= V * R1/ (R1+ R2) Voltage across R2= IR2= V * R2/ (R1+ R2)Î So voltage drops in proportion to resistance! Let V = 24, R1= 8Ω, R2= 4Ω V1= 24 * 8 / 12 = 16 V1= 24 * 4 / 12 = 8Î Useful when considering how resistors divide up voltagePHY2054: Chapter 1822CircuitsÎConsider the network of resistors shown below. When the switch S is closed, then: What happens to the voltage across R1, R2, R3, R4? What happens to the current through R1, R2, R3, R4? What happens to the total power output of the battery? Let R1= R2= R3= R4= 90 Ω and V = 45 V. Find the current through each resistor before and after closing the switch.Before¾I1= 45/135 = 1/3¾I2= 0¾I3= I4= 15/90=1/6After¾I1= 45/120 = 3/8¾I2= I3= I4= 1/8PHY2054: Chapter 1823CircuitsÎThe light bulbs in the circuits below are identical. Which configuration produces more light? (a) circuit I (b) circuit II (c) both the sameCircuit II has ½ current of each branchof circuit I, so each bulb is ¼ as bright.The total light in circuit I is thus 4x thatof circuit II.PHY2054: Chapter 1824CircuitsÎThe 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 muchBranch of circuit A has ½ resistance,thus it has 2x current.PHY2054: Chapter 1825CircuitsÎThe 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 muchUse P = I2R. Thus 2x current in Ameans it is 4x brighter.PHY2054: Chapter 1826More Complicated CircuitsÎParallel and series rules are not enough!ÎUse Kirchoff’s rulesPHY2054: Chapter 1827Problem Solving Using Kirchhoff’s RulesÎLabel 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: ÎApply 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 loopÎSolve equations simultaneously You need as many equations as you have unknownsÎSee example next slideII1I2I = I1+ I2+V - IRIPHY2054: Chapter 18289 V22 Ω15 Ω6 VExample 1: Applying Kirchhoff’s Rules ÎDetermine magnitudes and directions of all currents Take two loops, 1 and 2, as shown Choose currents & directions!: I1, I2, I3I1I2I3Use I1= I2+ I3323615 022 9 15 0III+−=−++ =2312315/22 0.686/15 0.401.08IIIII=====+=1212I1PHY2054: Chapter 1829Example 2: Applying Kirchhoff’s Rules ¾ Pick currents as shown¾ I3= I1+ I2121310 6 14 4 010 6 2 0IIII+−++ =+−− =1212312231IIIII==−=+ =−PHY2054: Chapter 1830CircuitsÎWhich of the equations is valid for the circuit loop below? a) 2 − I1− 2I2= 0 b) 2 − 2I1− 2I2− 4I3= 0 c) 2 − I1− 4 − 2I2= 0 d) I3− 2I2− 4I3= 0 e) 2 − 2I1− 2I2− 4I3= 0PHY2054: Chapter 1831Circuit Problem (1)ÎThe 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 changesBefore: Potential at (a) is 12V, but sois potential at (b) because equalresistance divides 24V in half. When theswitch is closed, nothing will changesince (a) and (b) are still at same potential.(a)(b)PHY2054: Chapter 1832Circuit Problem (2)ÎThe 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 changesBefore: Potential at (a) is 12V, but sois potential at (b) because equalresistance divides 24V in half. When the switch is closed, nothing will change since (a) and (b) are still at same potential.(b)(a)PHY2054: Chapter 1833(a)(b)Circuit Problem (3)ÎThe 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 outBefore: Bulb A and bulb B both have 18V across them.After: Bulb A has 12V across it andbulb B has 24V across it (these potentials are forced by the
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