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Transcript
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 Suppose we have the following circuit: I1 10 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5V I3 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! At the cross points, Iin = Iout I1 10 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5V I3 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 5V I3 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. 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. 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. Solve for the currents: I1 10 V 5 Ohms 2 Ohms 3 Ohms I2 8 Ohms 6 Ohms 5V I3 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… 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 2nd. 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 3rd 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 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 Try a less difficult one: I1 10 V 5 Ohms 2 Ohms 4 Ohms I2 0 Ohms 0 Ohms 4V I3 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… 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 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.