1 / 26

Circuits, Currents, and Kirschoff’s Laws

Circuits, Currents, and Kirschoff’s Laws. A PROJECT BY JAMES SABO AND SALLY JUNE TRACY. VOLTAGE ELECTRIC POTENTIAL ENERGY DIFFERENCE PER UNIT CHARGE BETWEEN TWO POINTS VOLTS (joules/coulomb). CURRENT MOVING ELECTRIC CHARGE IS IN THE DIRECTION OF POSITIVE CHARGE FLOW.

mikkel
Download Presentation

Circuits, Currents, and Kirschoff’s Laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Circuits, Currents, and Kirschoff’s Laws A PROJECT BY JAMES SABO AND SALLY JUNE TRACY

  2. VOLTAGE ELECTRIC POTENTIAL ENERGY DIFFERENCE PER UNIT CHARGE BETWEEN TWO POINTS VOLTS (joules/coulomb) CURRENTMOVING ELECTRIC CHARGE IS IN THE DIRECTION OF POSITIVE CHARGE FLOW. NET RATE AT WHICH CHARGE CROSSES A GIVEN AREA DRIVEN BY VOLTAGE DIFFRENCE AMPERES (coulombs/second)

  3. RESISTANCE • In most conductors charge doesn’t move unimpeded • Charge looses energy gained from electric potential difference in collisions • This is described by the quantity of resistance specific to different conductors • In units of Ohms (volt meters)

  4. V = IR • The relationship between voltage current and resistance • The potential drop across a circuit element is linearly proportional to the current R1 I1

  5. V = IR in Practice • Lets suppose V1 = 6V and R1 = 3Ώ • V = IR  I = V/R • I = (6V/ 3Ώ) = 2A

  6. Kirschoff Loop Law • The sum of the voltages in a circuit is equal to zero • Across a resistor: • With the current becomes negative (voltage drop) • Against the current becomes positive (voltage gain) • V1 - I1R1 - I1R2 = 0 • Or • I1R1 + I1R2 = V1 R1 R2 V1

  7. Kirschoff Node Law • The sum of the currents into and out of a node are equal to zero. • Current in is positive • Current out is negative • I1 – I2 – I3 = 0 •  I1 = I2 + I3

  8. Kirschoff’s Laws in Action • As you can see, with multiple voltage sources and multiple paths, V = IR will not solve for the currents here. • Kirschoff’s laws must be applied. A

  9. Apply Node Law • At point A: I1 + I2 - I3 = 0 • Or I1 + I2 = I3 • If the current direction is not known, simply guess. • Based on the circuit, it looks like a good guess for the direction of each, but if we are wrong, don’t worry! The current will just become negative. A

  10. Apply Loop Law • You can pick up to three different loops here. The two shown and one could be picked around the perimeter. • From loop 1: • V1 - I1R1 - I3R2 = 0 • From loop 2: • V2 - I2R3 - I3R2 = 0 Loop 1 Loop 2

  11. Solving the Circuit • We have three equations and three unknowns, the circuit can now be solved. • WARNING, This section involves rigorous algebra, you may become lost, confused or bored in the process. • Just bear with us it will be over soon enough.

  12. From the three equations: 1. I1 + I2 = I3 I1 = I3 – I2 2. V1 - I1R1 - I3R2 = 0 3. V2 - I2R3 - I3R2 = 0 Sub 1. Into 2. V1 – (I3 – I2)R1 - I3R2 = 0 Solve for I3. 4. I3 = (V1 - I2R1)/(R1 + R2) Sub 4 into 3. V2 - I2R3 - [(V1 - I2R1)/(R1 + R2)]R2 = 0 Solve for I2 I2 =[V2 – (V1R2)/(R1 + R2)] / [(R3 – (R1R2)/(R1 + R2)] Use given values: I2 = 1.20A Solving the Circuit

  13. Plug I2 into 4. I3 = (V1 - I2R1)/(R1 + R2) I3 = 1.08A Sub I2 and I3 into 1 I1 = I3 – I2 I1 = -.13A Now the circuit is solved! As you can see that method involves a lot of algebra, and a lot of keeping track of which equation is where. Fortunately for us, there is a better way of doing things. Still Solving the Circuit

  14. Linear Systems…My Savior! • You have seen before how currents can be solved using rigorous algebra, but a much simpler method can be used. • Fortunately we are certain that anyone in this class can solve this circuit using basic row operations we have learned in this class.

  15. Go back to the three original equations. 1. I1 + I2 = I3 2. V1 - I1R1 - I3R2 = 0 3. V2 - I2R3 - I3R2 = 0 Put all currents on the left and all non currents on the right. I1 + I2 - I3 = 0 I1R1 + I3R2 = V1 I2R3 + I3R2 = V2 As you should be able to see, this can become a matrix. [ I1 + I2 - I3 : 0 ] [ I1R1 + I3R2 : V1] [ I2R3 + I3R2: V2] Use Linear Systems!

  16. An augmented matrix can be formed. [ 1 1 -1: 0 ] [ R1 0 R2 : V1] [ 0 R3 R2 : V2 ] Plug in the values [ 1 1 -1: 0 ] [ 3 0 5 : 5 ] [ 0 8 5 :15] Find the RREF of the matrix [ 1 0 0 : -.1265822] [ 0 1 0 : 1.202531 ] [ 0 0 1 : 1.075949] This corresponds to each of the currents. I1 = -.1265822A I2 = 1.202531A I3 = 1.075949A Taking the RREF

  17. The Mesh Method • A different form of Kirschoff’s loop laws to solve for currents is called the Mesh Method. • Instead of assigning a current through each resistor, assign a current through each loop. • Once the currents are found, add the currents flowing through each resistor.

  18. Mesh Method Application • Going back to the old circuit we used, instead of using three equations (two loop laws and a node law) It can be broken into two equations. • V1 - I1R1 - (I1 + I2)R2 = 0 • V2 - I2R3 - (I2 + I1)R2 = 0 Loop 1 Loop 2

  19. Mesh Method Application • The currents can then be solved for in each equation and put into a matrix. • I1(R1 + R2) + I2R2 = V1 • I1R2 + I2(R3 + R2) = V2 • [ (R1 + R2) R2 : V1 ] • [ R2 (R3 + R2) : V2 ]

  20. Mesh Method Application • Plugging in original values, currents can be solved. The currents through each resistor are then added to solve for each current. • [ (3Ώ + 5Ώ) 5Ώ : 5V ]  [ 8 5 : 5 ] • [ 5Ώ (8Ώ + 5Ώ) :15V]  [ 5 13 : 15] • RREF = [ 1 0 : -.1265822]  I1 = -.1265822A • [ 0 1 : 1.202531 ]  I2 = 1.202531A

  21. Solving for the Currents • We now know that I1 = -.1265822A and that I2 = 1.202531A • Therefore the current through R1 is I1= -.13A The current through R3 is I2 = 1.20A and the current through R2 is (I1 + I2) = 1.07A

  22. Why use the Mesh Method? • You may wonder why the mesh method can be helpful. In the last exercise, instead of 3 equations, there were 2. That doesn’t save too much time. • When the circuits start to get more tricky, the mesh method starts to look a heck of a lot more useful.

  23. Monster Circuit Six currents, which means six equations!

  24. Quickly Analyze the circuit • V3 - (I1 +I3)R3 - I1R6 - (I1-I2)R1 - V1 = 0 • V1 - (I2 - I1)R1 - I2R4 - (I2 + I3)R2 - V2 = 0 • V3 - (I1 + I3)R3 - I3R5 - (I2 + I3)R2 - V2 = 0 • I1(-R3 - R6 - R1) + I2(R1) + I3(-R3) =V1 - V3 • I1(R1) + I2(-R1 - R4 - R2) +I3(-R2) = V2 - V3 • I1(-R3) + I2(-R2) + I3(-R3 - R5 - R2) = V2 - V3

  25. Put it in a matrix and take the RREF • [-R3 - R6 - R1 R1 -R3 : V1 - V3 ] • [ R1 -R1 - R4 - R2 -R2 : V2 - V1 ] • [ -R3 -R2 -R3 - R5 - R2 : V2 - V3 ] • Note that: • i1 = I2 - I1; i2 = -I2 - I3; i3 = I1 + I3 • i4 = I2; i5 = I3; i6 = I1 + I3 • AND WE’RE DONE!

  26. Thank you for paying attention! • We hope you all have a wonderful day and that you have somewhat been enlightened in the ways of linear systems! • Goodbye

More Related