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Kirchhoff part 2

Kirchhoff part 2. Starter . Starter . 1.5 Ω. 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 . Kirchhoff’s Second Law. I am back!!

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Kirchhoff part 2

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  1. Kirchhoff part 2

  2. Starter

  3. Starter 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?

  7. 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.

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

  9. 6.0V 1Ω I 7Ω 4Ω A battery of e.m.f. 6.0V and internal resistance 1Ω is connected to two resistors of 4Ω and 7Ω in series: calculate The total resistance in the external circuit The current supplied by the battery The terminal p.d. of the battery

  10. A 12V car battery is recharged by passing a current through it in the reverse direction using a 14V charger. Calculate the charging current 14V 12V 0.040Ω 0.050Ω 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)

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