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Series Circuits

Series Circuits. AKA Voltage Dividers Cardinal Rules for Series Circuits

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Series Circuits

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  1. Series Circuits AKA Voltage Dividers Cardinal Rules for Series Circuits Potential difference is divided proportionately amongst each of the resistors placed in series with the battery. The sum of the drops in potential at each device in the circuit should add the theemf applied to the circuit. (Voltage is constant) Energy In = Energy Out E = V1 + V2 + … + Vn The total current drawn from the battery is the same through any resistor placed in series with the battery. IT = I1 = I2 = ….. = In The total resistance in a circuit is equal to the sum of all the individual resistances placed in series. RT = R1 + R2 + … + Rn

  2. Series Circuits A R1 r E R2 B R3

  3. Series Circuits A R1 r = E R2 + B + R3 + All resistances are added together to provide the RT

  4. Series Circuits A R1 r E R2 B R3 The total current drawn from the battery is determined by the total resistance.

  5. Series Circuits A = 0.148A A A = 0.148A R1 A r = E R2 = A = 0.148A B A = 0.148A = R3 The total current drawn from the battery flows through every part of this circuit. Iis constant =

  6. Series Circuits A R1 r E R2 B R3

  7. Series Circuits V = 2.22 V A R1 r V = 0.030 V E R2 x = V = 3.74 V B V = x R3 As electric current flows across each resistor it spends energy & those moving charges experience a drop in their potential (voltage drop) = x

  8. Series Circuits A R1 r E R2 B R3 V3 could be calculated the same way….. = x

  9. Series Circuits A R1 r - E R2 - B - R3 = …. Or V3 could be the remaining difference in potential.

  10. Series Circuits A R1 r VAB= E – IT r Less = VAB E R2 B VAB= 9.00 V – 0.030 V = 8.97 V R3 Terminal voltage or VAB is the energy supplied by the battery that is supplied to and used in the circuit.

  11. Parallel Circuits • AKA Current Dividers • Cardinal Rules for Parallel Circuits • Potential difference is the same across all parallel paths (Voltage is constant) • E = V1 = V2 = ….. = Vn • The total current drawn from the battery is divided proportionately amongst each of the parallel paths • IT = I1 + I2 + ….. + In • One resistor having the same resistance as all resistance offered by the load on the parallel paths is referred to as the equivalent resistance. A similar idea to total resistance in series circuits. The equivalent resistance is equal to the reciprocal of the sum of the reciprocals of those resistors in parallel. • Req = _______1__________ • 1/R1 + 1/R2 + … + 1/Rn

  12. Parallel Circuits R2 E R3 R1 These are the same 3 resistors just arranged in parallel.

  13. Parallel Circuits First, we need to solve for Req Req= _______1__________ 1/R1 + 1/R2 + 1/R3 = _________1_________ 1/15.0Ω + 1/25.25Ω + 1/20.50Ω = 6.45Ω NOTE: The value of Req is always smaller than the smallest resistor in parallel. R2 E R3 R1

  14. Parallel Circuits A = IT R2 E R3 R1 Using Ohm’s Law still calculates the total current drawn from the battery.

  15. Parallel Circuits R2 E R3 = V = E R1 = = The emf applied to the circuit is applied across each parallel branch. V is constant.

  16. Parallel Circuits A = I3 R2 A = I1 E R3 R1 A = I2 Using the voltage drops across each resistor & the resistance, current flowing across each of the resistors can be found. I is divided amongst the parallel branches.

  17. Parallel Circuits R2 E R3 R1 I3 can be found using the exact same method as the other currents or…..

  18. Parallel Circuits R2 E R3 - R1 - = …or by finding the remaining current. IT = I1 + I2 + I3

  19. Parallel Circuits A r R2 E R3 R1 B How would this change if we put the internal resistor back into the circuit?

  20. Parallel Circuits A r R2 E R3 R1 B Req The internal resistor is always treated as a resistor in series with the battery and the Req

  21. Parallel Circuits A r R2 E R3 R1 B The total resistance still determines the current drawn from the battery.

  22. Parallel Circuits A r R2 = E R3 R1 B Since the r is considered to be in series with the battery it receives the IT.

  23. Parallel Circuits A r R2 E R3 = x R1 B Ohm’s Law calculates the energy spent moving the current out of the battery (volatge drop).

  24. Parallel Circuits A r R2 E R3 R1 = B VAB = E – IT r = 9.00 V – 0.27 V = 8.73 V The VABis the energy actaully supplied to the circuit AND what is applied across each parallel branch.

  25. Parallel Circuits A r R2 E R3 R1 B The VABis the energy actaully supplied to the circuit AND what is applied across each parallel branch.

  26. Parallel Circuits A r R2 E R3 R1 B Ohm’s Law determines the portion of the total current flowing down each parallel branch.

  27. Parallel Circuits A r R2 E R3 R1 B There is an alternate solution for I3 as it is the remaining current.

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