kirchoff's rules
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This document discusses Kirchhoff's rules for calculating current and voltage in circuits. It explains that Kirchhoff's junction rule states that the sum of all currents entering a junction must equal the sum of all currents leaving. Kirchhoff's loop rule states that the algebraic sum of changes in potential around any closed circuit loop must be zero. The document then works through an example calculation using these rules to find the currents I1, I2, and I3 and the total voltage VT in a circuit with multiple branches and loops.
kirchoff's rules
- 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.
- 7. CALCULATING CURRENT: USING KIRCHHOFF’S RULES Consider the figure shown. Find: I1 =? I2 =? I3 =? 𝑽 𝑻 =? 1 2 3 4 5
- 8. Solution We begin by applying Kirchhoff’s first or junction rule at point a. This gives I1 = I2 + I3 back
- 9. The loop rule The loop rule states that the changes in potential sum to zero. Thus, −I2R2 + emf1 − I2r1 − I1R1 =0 a→b b→c c→d d→a −I2(R2 + r1) + emf1 − I1R1 = 0 explain
- 10. Substitute the values for loop abcdea −I2(R2 + r1) + emf1 − I1R1 = 0 −I2(2.5Ω + 0.5Ω) + 18v − I1 6.0 Ω= 0 −3I2 + 18 − 6I1 = 0 back
- 11. Substitute the values for loop aefgha Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives + I1R1 + I3R3 + I3r2 − emf2 = + I1R1 + I3(R3 + r2) − emf2 = 0 Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes +6I1 + 2I3 − 45 = 0 back
- 12. These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for I2: ◦ I2 = 6 − 2I1 Now solve the third equation for I3: ◦ I3 = 22.5 − 3I1 Substituting these two new equations into the first one allows us to find a value for I1: ◦ I1 = I2 + I3 = (6−2I1) + (22.5− 3I1) = 28.5 − 5I1. Combining terms gives 6I1 = 28.5, and I1 = 4.75 A
- 13. I 2 = 6 − 2I1 = 6 − 9.50 ◦ I2 = −3.50 A I 3 = 22.5 − 3I 1 = 22.5 − 14 . 25 ◦ I3 = 8.25 A
- 14. Checking… junction rule I1 = I2 + I3 4.75 A=−3.50 A+ 8.25 A 4.75 A= 4.75 A
- 15. Checking… loop rule in loop abcdea 𝑉 = 0 −3I2 + 18 − 6I1 = 0 -10.5+18-28.5=0 0=0
- 16. If the formula in computing voltage is, 𝑉 = 𝐼𝑅 then:
- 17. THANK YOU..!
- 18. Quiz Compute the voltage a.)𝑅1 b. )𝑅2 𝑐. )𝑅3. Refer to the figure used in discussion and the data obtained in the computation.
- 19. answer