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A deep philosophical question to consider. Do you always eat your food from the outside in?

Step-by-step guide on solving circuit problems and calculating current and equivalent resistance.

nicholasgardner
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A deep philosophical question to consider. Do you always eat your food from the outside in?

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  1. A deep philosophical question to consider. Do you always eat your food from the outside in?

  2. Example: For the circuit below, calculate the current drawn from the battery and the current in the 6  resistor. 10  8  6  3  8  1  9 V The next three slides show the strategy for solving this. I will work the example at the blackboard in lecture.

  3. In a few minutes, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.” 10  8  6  3  8  1  9 V Replace the parallel combination (green) by its equivalent. Do you see any bite-sized chunks that are simple series or parallel?

  4. Any more “bite-sized chunks?” Remember that everything inside the green box is equivalent to a single resistor. 10  8  6  3  8  1  9 V Replace the series combination (blue box) by its equivalent.

  5. We are left with an equivalent circuit of 3 resistors in series, which is easy handle. 10  8  6  3  8  1  9 V Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination (orange) by its equivalent.

  6. Let’s shrink the diagram a bit, and work this a step at a time. R1=10  R3 and R4 are in parallel. R3=8  R2=6  R5= 3  R4=8  R6=1  =9 V R1=10  R2 and R34 are in series. R2=6  R34=4  R5= 3  R6=1  =9 V

  7. Let’s shrink the diagram a bit, and work this a step at a time. R1=10  R1 and R234 are in parallel. R234=10  R5= 3  R6=1  =9 V R1234, R5 and R6 are in series. R1234=5  R5= 3  R6=1  =9 V

  8. Calculate the current drawn from the battery. R1234=5  R5= 3  R6=1  =9 V I=1 A

  9. Find the current in the 6  resistor. R1=10  R3=8  There are many ways to do the calculation. This is just one. R2=6  R5= 3  R4=8  R6=1  I=1 A =9 V R1=10  V1 = V234 = V1234 (parallel). R234=10  R5= 3  I=1 A R6=1  =9 V

  10. Find the current in the 6  resistor. I V1234 = I R1234 = (1)(5) = 5 V R1234=5  R5= 3  V1 = V234 = 5 V I=1 A R6=1  =9 V R1=10  V234 = I234 R234 I234 I234 = V234 / R234 = 5/10 R234=10  R5= 3  I234 = 0.5 A I=1 A R6=1  =9 V

  11. Find the current in the 6  resistor. R1=10  R3=8  I2=I234 I234 = I2 = I34 = 0.5 A R2=6  R5= 3  I2 = 0.5 A R4=8  R6=1  I=1 A =9 V

  12. Find the current in the 6  resistor. R1=10  A student who has taken a circuits class will probably say R234=10  R5= 3  R1 = R234 so I1 = I234 = I/2 = 0.5 A R6=1  I=1 A =9 V If you want to do this on the exam, make sure you write down your justification on the exam paper, and don’t make a mistake! If you don’t show work and make a mistake, we can’t give partial credit. Answers without work shown generally receive no credit.

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