1. Lecture 91 Single Node-Pair Circuits and Current Division

2. Lecture 92 I R1R1 R2R2 V + - I1I1 I2I2 Example: 2 Light Bulbs in Parallel How do we find I 1 and I 2 ?

3. Lecture 93 Apply KCL at the Top Node I 1 + I 2 = I

4. Lecture 94 Solve for V

5. Lecture 95 Equivalent Resistance If we wish to replace the two parallel resistors with a single resistor whose voltage-current relationship is the same, the equivalent resistor has a value of:

6. Lecture 96 Now to find I 1 This is the current divider formula. It tells us how to divide the current through parallel resistors.

7. Lecture 97 What is the formula for I 2 ?

8. Lecture 98 1.17A 144  V + - I1I1 I2I2 Example: 2 Light Bulbs in Parallel Find I 1 and I 2 360 

9. Lecture 99 I 1 and I 2

10. Lecture 910 Example: 3 Light Bulbs in Parallel How do we find I 1, I 2, and I 3 ? I R2R2 V + - R1R1 I1I1 I2I2 R3R3 I3I3

11. Lecture 911 Apply KCL at the Top Node I 1 + I 2 + I 3 = I

12. Lecture 912 Solve for V

13. Lecture 913 R eq

14. Lecture 914 No Current Divider! We cannot make a simple current divider equation for three or more parallel resistors. We have to solve for V, then solve for the current(s) of interest.

15. Lecture 915 Example: 3 Light Bulbs in Parallel Find I 1 1.67A V + - I1I1 I2I2 I3I3 144  360  240 

16. Lecture 916 Compute V, then I 1

17. Lecture 917 Is2Is2 V R1R1 R2R2 + - I1I1 I2I2 More Than One Source How do we find I 1 or I 2 ? Is1Is1

18. Lecture 918 Apply KCL at the Top Node I 1 + I 2 = I s1 - I s2

19. Lecture 919 Multiple Current Sources We find an equivalent current source by algebraically summing current sources. We find an equivalent resistance. We find V as equivalent I times equivalent R. We then find any necessary currents using Ohm’s law.