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Thomas Oberst Cornell University Electrical Circuits.

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Presentation on theme: "Thomas Oberst Cornell University Electrical Circuits."— Presentation transcript:

1 Thomas Oberst Cornell University Electrical Circuits

2 Thomas Oberst Cornell University How you should be thinking about electric circuits: Voltage: a force that pushes and pulls the current through the circuit (in this picture it would be equivalent to gravity)

3 Thomas Oberst Cornell University Resistance: friction that impedes flow of current through the circuit (rocks in the river) How you should be thinking about electric circuits:

4 Thomas Oberst Cornell University Current: the actual “substance” that is flowing through the wires of the circuit (electrons!) How you should be thinking about electric circuits:

5 Thomas Oberst Cornell University Sample Circuit: Measured resistance of each light bulb to be 11 Ohms 12V 11  Will apply a fixed voltage of 12V across the terminals

6 Thomas Oberst Cornell University When I connect the 12V, which light bulb will light the brightest? Which light bulb will light the dimmest? Rate the bulbs in order from brightest to dimmest?

7 What is the resistance of each branch of the circuit? Branch 1 (bulbs 2 & 5): Branch 2 (bulbs 1, 3, & 4): R = = 22  R = /(1/11 + 1/11) = /(2/11) = /2 = = 16.5  Let’s find out if you’re guess is correct:

8 Thomas Oberst Cornell University What is the current flowing through each branch of the circuit? Branch 1 (bulbs 2 & 5): Branch 2 (bulbs 1, 3, & 4): V = IR I = V/R = 12/22 = 0.54 A I = V/R = 12/16.5 = 0.73 A

9 Thomas Oberst Cornell University Therefore, what must be the TOTAL current flowing through the circuit? I = I 1 + I 2 = = 1.27 A However, if I actually measure the current using an Ammeter, I get… I = 0.15 A

10 Thomas Oberst Cornell University That would mean that the light bulbs each have a resistance of: V tot = I tot R tot 12 = 0.15 [ 1/(1/(R branch 1 ) + 1/(R branch 2 )) ] = 0.15 [ 1/(1/(R+R) + 1/(R + 1/(1/R + 1/R))) ] = 0.15 [ 1/(1/2R + 1/(R + 1/(2/R))) ] = 0.15 [ 1/(1/2R + 1/(R + R/2)) ] = 0.15 [ 1/(1/2R + 1/(3R/2)) ] = 0.15 [6R/7] Or, R = (12 x 7)/(0.15 x 6) = 93  = 0.15 [ 1/(1/2R + 2/3R)]

11 Thomas Oberst Cornell University What the heck is going on?

12 Thomas Oberst Cornell University When I measured the resistance of the light bulbs, what temperature were they at? Did the temperature of the bulbs change when I turned on the 12V power supply? Is resistance dependant on temperature?

13 Thomas Oberst Cornell University In fact, resistance is related to temperature by the equation: R(T) = R(T 0 ) [1+ a  T ] Where  T = T – T 0 is the difference between two temperatures And a is the “temperature coefficient of resistivity” and depends on the material that the light bulb filaments are made of. All regular light bulbs have tungsten filaments, with a = o C -1

14 Thomas Oberst Cornell University Recall that I measured R = 11  at room temperature and I calculated that R must be 93  when the bulbs are lit (we will assume, for simplicity, that all the bulbs are the same temperature, which is not true because they are lit at different brightnesses). Can we use this information to actually calculate the temperature of the light bulbs?

15 Thomas Oberst Cornell University R(T) = R(T 0 ) [ 1 + a  T] 93 = 11 [ (T high – 23)] T high = [ (93/11 – 1) / ] + 23 T high = 1679 o C This is the temperature of the filament, not of the glass bulb surrounding the filament!

16 Thomas Oberst Cornell University Now that we know the correct resistance (93  ) of each bulb, back to figuring out why #1 was the brightest…

17 Thomas Oberst Cornell University What is the resistance of each branch of the circuit? Branch 1 (bulbs 2 & 5): Branch 2 (bulbs 1, 3, & 4): R = = 22  R = /(1/11 + 1/11) = /(2/11) = /2 = = 16.5  R = = 186  R = /(1/93 + 1/93) = /(2/93) = /2 = = 

18 Thomas Oberst Cornell University What is the current flowing through each branch of the circuit? Branch 1 (bulbs 2 & 5): Branch 2 (bulbs 1, 3, & 4): V = IR I = V/R = 12/22 = 0.54 A I = V/R = 12/16.5 = 0.73 A V = IR I = V/R = 12/186 = 0.065A I = V/R = 12/139.5 = A

19 Thomas Oberst Cornell University Therefore, what must be the TOTAL current flowing through the circuit? I = I 1 + I 2 = = 1.27 A However, if I actually measure the current using an Ammeter, I get… I = 0.15 A I = I 1 + I 2 = = 0.15 A

20 Thomas Oberst Cornell University So we know the total current, and the current flowing through each branch… Now what is the current flowing through each light bulb? #2? #5? #4? #3? #1? A A A

21 Thomas Oberst Cornell University What happens if I remove bulb #4? Is the current going through all four bulbs the same? What is it? A 2 x = 0.13 A What is it? Is the total current still the same?

22 Thomas Oberst Cornell University What happens if I remove bulb #5 (after replacing bulb #4)? Why did branch 1 go out? Why didn’t branch 2 get brighter?!?! (i.e., what happened to the current that was previously going through branch 1?)

23 Thomas Oberst Cornell University To understand the answer, you must realize that we are holding the total voltage fixed at 12V, and by taking out bulbs #2 and #5 I changed the total resistance.

24 Thomas Oberst Cornell University So with V fixed and R being changed, I HAS to change (recall V=IR). So it does not remain 0.15 A as it has been up until now. Therefore, the current from branch 1 doesn’t need to “go” anywhere. The total current of the whole circuit has simply dropped. (By an amount exactly equal to the current that used to be in branch 1).

25 Thomas Oberst Cornell University The End


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