PHY2054: Chapter 1821CircuitsÎThe 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 1822CircuitsÎThe 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 1823CircuitsÎThe 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 1824More Complicated CircuitsÎParallel and series rules are not enough!ÎUse Kirchoff’s rulesPHY2054: Chapter 1825Problem 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 - IRPHY2054: Chapter 18269 V22 Ω15 Ω6 VExample 1: Applying Kirchhoff’s Rules ÎDetermine magnitudes and directions of all currents Take two loops, 1 and 2, as shown Define currents: I1, I2, I3I1I2I3Use I1= I2+ I3323615 022 9 15 0III+−=−++ =2312315/22 0.686/15 0.401.08IIIII=====+=1212I1PHY2054: Chapter 1827Example 2: Applying Kirchhoff’s Rules Use I3= I1+ I2121310 6 14 4 010 6 2 0IIII+−++ =+−− =1212312231IIIII==−=+ =−PHY2054: Chapter 1828CircuitsÎWhich 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 1829Circuit 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 1830Circuit 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 1831(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 batteries).24VPHY2054: Chapter 1832Wheatstone BridgeÎAn ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. What is the current through the ammeter? a) I / 2 b) I / 4 c) zero d) need more informationThe parallel branches have the sameresistance, so equal currents flow ineach branch. Thus (a) and (b) are atthe same potential and there is nocurrent flow across the
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