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Published byIyanna Brabazon Modified about 1 year ago

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Goal: To understand how to solve circuits with multiple power sources Objectives: 1) To learn Kirchhoff’s Laws 2) To use those laws in order to play Physics Chutes & Ladders

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Suppose we have the following circuit: I110 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5 V I3

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Kirchhoff’s Laws You have already learned them! Law 1: The current going into a point must be the current going out of the point. So Iin = Iout Law 2: If you do a full loop, the net change in voltage is 0. We can use this to our advantage!

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At the cross points, Iin = Iout I110 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5 V I3

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If we do any loops then the net voltage is 0. That is the up voltage = the down voltage I1 10 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5 V I3

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Physics Chutes and Ladders When you look at a circuit, first plot the potential raises and drops. In the direction of the current for each section, the voltages for from low (-) to high (+) potentials. Resistors go from + to – What direction should the current go? Well don’t worry about it. If you choose wrong you will just get a negative current, and the is okay. Just be consistant.

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To solve the circuit: In this example you have 4 equations. The first is the easiest. Find a point where the 3 currents meet. Iin = Iout is the equation. The next 3 are similar.

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Current Loops Each way you can go in a full circle is a loop. In this example there are 3 loops (full perimeter, bottom circle, top circle). For each pick a direction (either one). On the left side of the equation you will write in the value of the voltage when you go UP. For a power source this is some # of volts. For a resister the voltage is IR. Be sure to include the correct value of I. Also, note that you can’t add up all the resistances in the loop because they have different currents. If any have the same current you can add those, but only if they are the same. On the right side of the equation you put in all the downs. Do this 3 times and you have a total of 4 equations.

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Solve for the currents: I110 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5 V I3

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Equations I1 = I2 + I3 Top: 10 V = 7 Ohms I1 + 3 Ohms I2 Bottom: 5V + 3 Ohms I2 = 14 Ohms I3 Perimeter: 5V + 10V = 7 Ohms I1 + 14 Ohms I3 Now we solve… Easiest to solve for 1 value in 3 different equations. Then you can get rid of it. Lets do it for I1…

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Equations I1 = I2 + I3 Top: I1 = 10/7 A - 3/7 I2 Perimeter: I1 = 15/7 A - 2 I3 Now we substitute I1 from first equation into the 2 nd. Then we solve for say I2. I2 + I3 = 10/7 A – 3/7 I2 I2 = 1 A – 0.7 I3 Okay, now we substitute I1 and I2 into the 3 rd equation for I1… I1 = I2 + I3 = 1 A + 0.3 I3 1A + 0.3 I3 = 15/7 A – 2 I3 So, 2.3 I3 = 8/7 A Or I3 = 0.50 A

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Finish Up Now we just solve for the rest. I1 = I2 + I3 Top: I1 = 10/7 A - 3/7 I2 Perimeter: I1 = 15/7 A - 2 I3 I3 = 0.5 A So, I1 = 15/7 A – 2 * 0.5 A I1 = 1.15 A 1.15 A = I2 + 0.50 A I2 = 0.65 A

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Try a less difficult one: I110 V 5 Ohms 2 Ohms 4 Ohms I2 0 Ohms 0 Ohms 4 V I3

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Set it up I1 = I2 + I3 Top: 10 V = 7 Ohms I1 + 4 Ohms I2 Bottom: 4V + 4 Ohms I2 = 0 Perimeter: 4V + 10V = 7 Ohms I1 This one can we worked out with a lot less steps…

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Set it up I1 = I2 + I3 Top: 10 V = 7 Ohms I1 + 4 Ohms I2 Bottom: 4V + 4 Ohms I2 = 0 Perimeter: 4V + 10V = 7 Ohms I1 This one is a lot easier… I2 = -1 A I1 = 2 A I3 = I1 – I2 = 3A

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Conclusion: Current in = Current out The net voltage around a loop is 0 so therefore the ups = the downs. This allows us to play the most boring game EVER invented: Physics Chutes and Ladders By using this to create our equations for the loops we are able to solve the circuit.

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