1. Do now: find the missing values 1) 2) 3) 4)

2. We have covered: LO1) Recall the basics of Year 12 electricity Today we are learning to: LO2) Explain Kirchhoff’s laws LO3) Use Kirchoff’s laws… to calculate voltage and current… in networks of series and parallel circuits Learning Outcomes: Rātapu, 6 Hakihea 2015

3. Gustav Robert Kirchhoff German physicist (1824 – 1887). He formulated his circuit laws in 1845, while still an undergraduate; it later became his PhD dissertation He proposed his law of thermal radiation in 1859 and coined the term “black body radiation” He discovered caesium and rubidium in 1861 with Robert Bunsen Kirchhoff (left) and Bunsen (right)

4. Kirchhoff’s Laws Current Law: The total current entering a junction in a circuit equals the total current leaving. I 1 = 2A I 2 = ? I 3 = 5A

5. Kirchhoff’s Laws Voltage Law: The total of all the potential differences around a closed loop in a circuit is zero V1V1 V3V3 V2V2 + –

6. Kirchhoff’s Laws For voltage sources: Going from – to + represents a gain in energy so potential difference is positive Going from – to + represents a gain in energy so potential difference is positive Going from + to – represents a loss of energy so potential difference is negative Going from + to – represents a loss of energy so potential difference is negative +V -V

7. Kirchhoff’s Laws For resistors: Passing in the same direction as the current represents a loss of energy -V=-(IR) Passing in the same direction as the current represents a loss of energy -V=-(IR) Passing in the opposite direction to the current represents a gain in energy +V=+(IR) Passing in the opposite direction to the current represents a gain in energy +V=+(IR) II -V = -IR +V = +IR

8. Kirchhoff’s Laws To solve problems, follow a voltage loop around a circuit… a-b a-b b-c b-c c-d c-d d-a d-a Then add up the p.d’s. V? 6V 4Ω4Ω 1Ω1Ω ab cd 2A 3A 0 +6 = -2A x 4Ω -8 +3 – V 0 + 6 – 8 + 3 – V = 0 V = 1V !

9. Kirchhoff’s Laws Harder…. a-b a-b b-c b-c c-d c-d d-a d-a Adding: Adding: So I 1 So I 1 Using the current law: 4Ω4Ω 2Ω2Ω 8Ω8Ω 5V V? 0.75A I1?I1? I2?I2? a b c d e f 0 -3 = -0.75A x 4Ω = -0.75A x 4Ω -I 1 x 2 +5 -3 -2I 1 +5 = 0 = 1A 1 = 0.75 + I 2 1 = 0.75 + I 2 So I 2 = 0.25A

10. Kirchhoff’s Laws Continued… b-c b-c c-f c-f f-e f-e e-b e-b Adding: Adding: 4Ω4Ω 2Ω2Ω 8Ω8Ω 5V V? 0.75A I1?I1? I2?I2? a b c d e f -3 = -0.75A x 4Ω +2 = +0.25A x 8Ω +V? 0 -3 + 2 + V = 0 So V = 1V