kirchoff's rules

Calculate the current and voltage through and across circuit elements using Kirchhoff’s loop and junction rules (at most 2 loops only)

1. KIRCHHOFF’S RULES Prepared by: RG Luis Vincent P. Gonzaga

2. Objective Calculate the current and voltage through and across circuit elements using Kirchhoff’s loop and junction rules (at most 2 loops only)

3. Kirchhoff’s first rule— the junction rule . The sum of all currents entering a junction must equal the sum of all currents leaving the junction .

4. Kirchhoff’s second rule— the loop rule . The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

5. A junction in a circuit is a point where three or more conductors meet.

6. Things to learn

7. CALCULATING CURRENT: USING KIRCHHOFF’S RULES Consider the figure shown. Find: I 1  =?   I 2   =?   I 3   =?   1 2 3 4 5

8. Solution We begin by applying Kirchhoff’s first or junction rule at point a . This gives I 1   =  I 2  +  I 3 back

9. The loop rule The loop rule states that the changes in potential sum to zero . Thus , − I 2 R 2  + emf 1  −  I 2 r 1  −  I 1 R 1  =0 a b b c c d d a − I 2 ( R 2  +  r 1 ) + emf 1  −  I 1 R 1  = 0   explain

10. Substitute the values for loop abcdea − I 2 ( R 2  +  r 1 ) + emf 1  −  I 1 R 1  = − I 2 (2.5 Ω   +  0.5 Ω ) + 18v   −  I 1   6.0 Ω = −3 I 2   + 18 − 6 I 1   = back

11. Substitute the values for loop aefgha Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives +  I 1 R 1  +  I 3 R 3  +  I 3 r 2  − emf 2  = +  I 1 R 1  +  I 3 ( R 3  +  r 2 ) − emf 2  = Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes +6 I 1  + 2 I 3  − 45 =  back

12. These three equations are sufficient to solve for the three unknown currents . First, solve the second equation for  I 2 : I 2  = 6 − 2 I 1 Now solve the third equation for  I 3 : I 3  = 22.5 − 3 I 1 Substituting these two new equations into the first one allows us to find a value for  I 1 : I 1  =  I 2  +  I 3   = (6−2 I 1 ) + (22.5− 3 I 1 ) = 28.5 − 5 I 1 . Combining terms gives 6 I 1  = 28.5, and I 1  = 4.75 A

13. I  2  = 6 − 2 I 1  = 6 − 9.50 I 2  = −3.50 A I  3  = 22.5 − 3 I  1  = 22.5 − 14 . 25 I 3  = 8.25 A

14. Checking… junction rule I 1  =  I 2  +  I 3 4.75 A= −3.50 A + 8.25 A 4.75 A= 4.75 A

15. Checking… loop rule in loop abcdea − 3 I 2   + 18 − 6 I 1   = -10.5+18-28.5=0 0=0  

16. If the formula in computing voltage is, then:  

17. Thank you..!

18. Quiz Compute the voltage a.) . Refer to the figure used in discussion and the data obtained in the computation.  

19. answer