1. Kirchhoff part 2

2. Starter

3. 1.5Ω

4. Learning objectives State Kirchhoff’s second law Apply Kirchhoff’s second law to circuits Solve circuit problems involving series and parallel circuits with one or more sources of e.m.f.

5. Kirchhoff’s Second Law I am back!! This time it is slightly more complicated

6. What’s it all about?

10. Kirchhoff’s second law The sum of the e.m.f.s is equal to the sum of the p.d.s in a closed loop. This is an example of conservation of energy. What it essentially says is that all the energy put into a circuit from the battery has to go somewhere. And this balance must be exact. You can't have even a small amount of energy appearing from nowhere or disappearing without trace.

14. V R r Є V I Terminal p.d. measured here Lost volts across the internal resistance e.m.f. = terminal p.d. + lost volts Є = V + v We know that current through both resistors is I So applying V= IR to each resistor Є = V +IrЄ = IR + Ir Є = I(R+r)

15. 6.0V 1Ω1Ω 7Ω7Ω 4Ω4Ω I A battery of e.m.f. 6.0V and internal resistance 1Ω is connected to two resistors of 4Ω and 7Ω in series: calculate a.The total resistance in the external circuit b.The current supplied by the battery c.The terminal p.d. of the battery

16. 12V 14V 0.040Ω 0.050Ω A 12V car battery is recharged by passing a current through it in the reverse direction using a 14V charger. Calculate the charging current Note the 12 V e.m.f. opposes the 14V e.m.f. so: the sum of the e.m.f.s = 14 + - 12 = 2V There are 2 internal resistors, and both ‘waste’ electrical energy Є = I( R +r)