nagle_phys2020_fa07_lecture19

nagle_phys2020_fa07_lecture19 - Clicker Question In the...

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Unformatted text preview: Clicker Question In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out, its resistance becomes infinite.) 2 A) Bulb 1 gets brighter B) Bulb 1 gets dimmer. C) It's brightness remains the same. 3 1 (Hint: What happens to the current from the battery when bulb 2 burns out.) V=12V Answers: Bulb 1 gets dimmer! When bulb 2 burns outs, the filament inside breaks and R2 becomes infinitely large. The total equivalent resistance which the battery sees increases (since bulb 2 is gone, there are fewer paths for the current flow, so less flow, more total resistance.) Since the battery sees a larger R-tot, the current from the battery I-tot = V/Rtot is reduced. Less current from the battery means less current 2 through bulb 1, less light. 1 3 V=12V 50 51 It is good to be able to think through (without lots of equations) how the components of these circuits work. However, we will need to use our equations to work out detailed values for current and electric potential in more complicated circuits. We now outline a systematic procedure for analyzing circuits. Example 1: Choose R1 = R2 = R3 = 10 1. Longer line on battery is higher voltage and the shorter line is lower voltage, i1 by convention. 2. We label the current in 1 each section as shown. Note that we can guess the ib direction at this point. The answer will be independent of our guess. ib 2 i2 i3 3 ib + V=12V ib 52 53 3. Apply the Voltage Loop Rule around each possible loop in the circuit. Make sure to label the direction of your loop! Make sure to pick a starting point to go around the loop. ib Then add up the Voltage changes around the loop. ib 1 Critical Sign Convention! 2 i2 - + V=12V i1 3 i3 - If your loop goes through a battery from to + the Voltage increases (e.g. V = +12 V) ib + V=12V ib + V=12V If your loop goes through a battery from + to the Voltage decreases (e.g. V = -12 V) 54 55 1 Critical Sign Convention! i 2 Starting at Point P go around the loop. If you go across a resistor and the loop direction and guessed current direction are the same, the voltage decreases (e.g. V = -iR) If you go across a resistor and the loop direction and guessed current direction are opposite, the voltage increases (e.g. V = +iR) Vbat = +12V VR 3 = -i3 R3 VR1 = -i1 R1 Vsum = +12 - i3 R3 - i1 R1 ib = 2 i2 i1 1 3 i3 i 2 0 P ib + ib V=12V ib 56 57 Now try other path starting at Point P around the new loop. 2 i2 For our specific circuit, we used R1 = R2 = R3 = 10 Vbat = +12V VR 2 = -i2 R2 VR1 = -i1 R1 Vsum = +12 - i3 R3 - i1 R1 = 0 Vsum = +12 - i2 R2 - i1 R1 = 0 ib 12 - 10i3 - 10i1 = 0 12 - 10i2 - 10i1 = 0 i1 1 3 i3 We have two equations and three unknowns. We need additional information. Vsum = +12 - i2 R2 - i1 R1 = 0 P ib ib + V=12V 58 59 Apply the current junction rule at point Q below. iin = iout i2 + i3 = i1 Also, (as we could have done at the start): 1 12 - 10i3 - 10i1 = 0 2 i2 12 - 10i2 - 10i1 = 0 i1 Q ib ib + V=12V 3 i3 i2 + i3 = i1 Solve 3 equations and 3 unknowns. ib i1 = 0.8 Amps i2 = 0.4 Amps i3 = 0.4 Amps i1 = ib ib 60 61 2 Does this make sense? i2 = i3 since two parallel, equal resistance paths. Current from R2 and R3 must go into R1. Clicker Question 2 i2 R1 = 10 i1 = 0.8 Amps 2 i2 i1 1 3 i3 i1 = 0.8 Amps i2 = 0.4 Amps i3 = 0.4 Amps i1 i1 i1 + V=12V i1 R2 = 10 i2 = 0.4 Amps What is the value of the equivalent resistor that R3 = 10 i3 = 0.4 Amps i1 would replace the three resistors below? A) Req = 5 1 B) Req = 10 i1 C)Req = 15 D)Req = 30 E) None of the Above i1 i3 3 i1 + V=12V i1 62 63 Two ways to think about it. 1. Battery of 12 V has 0.8 Amps coming out of it. By V=iR the Req = 12V/0.8A=15 2. R2 and R3 are in parallel. R23 = 1 1 1 R +R 3 2 = 5 R1 = 10 i1 = 0.8 Amps R2 = 10 i2 = 0.4 Amps R3 = 10 i3 = 0.4 Amps Clicker Question 2 i2 Which of these diagrams represent the same circuit? i1 1 3 i3 A. a and b B. a and c C. b and c D. a, b, and c E. a, b, and d i1 i1 + Then R23 and R1 are in series. i1 V=12V i1 Req = R123 = R1 + R23 = 15 64 65 Clicker Question A circuit with two batteries is shown below. The directions of the currents have been chosen (guessed) as shown. Which is the correct current equation for this circuit? A) I2 = I1 + I3 B) I1 = I2 + I3 C) I3 = I1 + I2 I1 D) None of these. R1 V1 R2 I2 Loop 1. R3 I3 V2 Clicker Question A circuit with two batteries is shown below. The directions of the currents have been chosen (guessed) as shown. Which is the correct current equation for this circuit? A) I2 = I1 + I3 B) I1 = I2 + I3 C) I3 = I1 = I2 I1 D) None of these. R1 V1 R2 I2 Loop 1. R3 I3 V2 66 67 3 Clicker Question Which equation below is the correct equation for Loop 1? B) V2 + I1R1 I2R2 = 0 A) V2 + I1R1 I2R2 = 0 C) V2 I1R1 + I2R2 = 0 D) V2 + I1R1 + I2R2 = 0 E) None of these. I1 I3 Answer: V2 + I1R1 I2R2 = 0 R1 V1 R2 I2 Loop 1. R3 I3 V2 Finish a full analysis of this circuit. Consider case: V1 = 12 V, V2 = 24 V, R1=R2=R3 = 1 . I3 I1 R1 V1 R2 V2 I2 Loop 1. Loop 2 R3 I3 i1 = 20 A, i2 = -4 A, i3 = 16 A i3 = i1 + i2 - V2 + i1 R1 - i2 R2 = 0 - 24 + i1 - i2 = 0 + V1 - i2 R2 - i3 R3 = 0 + 12 - i2 - i3 = 0 68 69 4 ...
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This note was uploaded on 02/27/2008 for the course PHYS 2020 taught by Professor Dubson during the Fall '06 term at Colorado.

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