Series Circuits AKA Voltage Dividers Cardinal Rules for Series Circuits Potential difference is divided proportionately amongst each of the resistors placed.

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Presentation transcript:

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 the emf applied to the circuit. (Voltage is constant) Energy In = Energy Out E = V 1 + V 2 + … + V n The total current drawn from the battery is the same through any resistor placed in series with the battery. I T = I 1 = I 2 = ….. = I n The total resistance in a circuit is equal to the sum of all the individual resistances placed in series. R T = R 1 + R 2 + … + R n

VIR Source Totals 9.00 V r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Series Circuits r R1R1 R2R2 R3R3 B A E

VIR Source Totals 9.00 V60.95Ω r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Series Circuits r R1R1 R2R2 R3R3 B A All resistances are added together to provide the R T = E

VIR Source Totals 9.00 V I T = E /R T A 60.95Ω r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Series Circuits r R1R1 R2R2 R3R3 B A E The total current drawn from the battery is determined by the total resistance.

VIR Source Totals 9.00 V0.148 A60.95Ω r A0.20Ω Resistor A15.00Ω Resistor A25.25Ω Resistor A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E The total current drawn from the battery flows through every part of this circuit. I is constant = = = = A = 0.148A A

VIR Source Totals 9.00 V0.148 A60.95Ω r A0.20Ω Resistor A15.00Ω Resistor A25.25Ω Resistor A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E

VIR Source Totals 9.00 V0.148 A60.95Ω r V0.148 A0.20Ω Resistor V0.148 A15.00Ω Resistor V0.148 A25.25Ω Resistor A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E As electric current flows across each resistor it spends energy & those moving charges experience a drop in their potential (voltage drop) V = V V = 3.74 V V V = 2.22 V = = = x x x

VIR Source Totals 9.00 V0.148 A60.95Ω r V0.148 A0.20Ω Resistor V0.148 A15.00Ω Resistor V0.148 A25.25Ω Resistor V0.148 A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E V 3 could be calculated the same way….. =x

VIR Source Totals 9.00 V0.148 A60.95Ω r V0.148 A0.20Ω Resistor V0.148 A15.00Ω Resistor V0.148 A25.25Ω Resistor V0.148 A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E …. Or V 3 could be the remaining difference in potential. = - - -

VIR Source Totals 9.00 V0.148 A60.95Ω r V0.148 A0.20Ω Resistor V0.148 A15.00Ω Resistor V0.148 A25.25Ω Resistor V0.148 A20.50Ω Series Circuits r R1R1 R2R2 R3R3 B A E Terminal voltage or V AB is the energy supplied by the battery that is supplied to and used in the circuit. V AB = E – I T r V AB = 9.00 V – V = 8.97 V Less = V AB

Parallel Circuits AKA Current Dividers Cardinal Rules for Parallel Circuits Potential difference is the same across all parallel paths (Voltage is constant) E = V 1 = V 2 = ….. = V n The total current drawn from the battery is divided proportionately amongst each of the parallel paths I T = I 1 + I 2 + ….. + I n 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

VIR Source Totals 9.00 V Resistor Ω Resistor Ω Resistor Ω Parallel Circuits R1R1 R2R2 R3R3 E These are the same 3 resistors just arranged in parallel.

Parallel Circuits R1R1 R2R2 R3R3 E 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 = 6.45 Ω NOTE: The value of Req is always smaller than the smallest resistor in parallel.

VIR Source Totals 9.00 V I T = E /Req 1.40 A 6.45Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits R1R1 R2R2 R3R3 E Using Ohm’s Law still calculates the total current drawn from the battery. A = I T

VIR Source Totals 9.00 V1.40 A6.45Ω Resistor V15.00Ω Resistor V25.25Ω Resistor V20.50Ω Parallel Circuits R1R1 R2R2 R3R3 E = = = The emf applied to the circuit is applied across each parallel branch. V is constant. V = E

VIR Source Totals 9.00 V1.40 A6.45Ω Resistor V I 1 = V 1 /R 1 = A 15.00Ω Resistor V I 2 = V 2 /R 2 = A 25.25Ω Resistor V20.50Ω Parallel Circuits R1R1 R2R2 R3R3 E 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. A = I 1 A = I 2 A = I 3

VIR Source Totals 9.00 V1.40 A6.45Ω Resistor V I 1 = V 1 /R 1 = A 15.00Ω Resistor V I 2 = V 2 /R 2 = A 25.25Ω Resistor V I 3 = V 3 /R 3 = A 20.50Ω Parallel Circuits R1R1 R2R2 R3R3 E I 3 can be found using the exact same method as the other currents or…..

VIR Source Totals 9.00 V1.40 A6.45Ω Resistor V A15.00Ω Resistor V A25.25Ω Resistor V0.444 A20.50Ω Parallel Circuits R1R1 R2R2 R3R3 E …or by finding the remaining current. I T = I 1 + I 2 + I 3 = - -

VIR Source Totals 9.00 V r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E How would this change if we put the internal resistor back into the circuit?

VIR Source Totals 9.00 V R T = Req + r = 6.45Ω+0.20Ω = 6.65Ω r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E The internal resistor is always treated as a resistor in series with the battery and the Req Req

VIR Source Totals 9.00 V I T = E /Req 1.35 A 6.65Ω r 0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E The total resistance still determines the current drawn from the battery.

VIR Source Totals 9.00 V1.35 A6.65Ω r 1.35 A0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E Since the r is considered to be in series with the battery it receives the I T. =

VIR Source Totals 9.00 V1.35 A6.65Ω r V r = I T x r 0.27 V 1.35 A0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E Ohm’s Law calculates the energy spent moving the current out of the battery (volatge drop). =x

VIR Source Totals 9.00 V1.35 A6.65Ω r 0.27 V1.35 A0.20Ω Resistor Ω Resistor Ω Resistor Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E The V AB is the energy actaully supplied to the circuit AND what is applied across each parallel branch. = V AB = E – I T r = 9.00 V – 0.27 V = 8.73 V

VIR Source Totals 9.00 V1.35 A6.65Ω r 0.27 V1.35 A0.20Ω Resistor V15.00Ω Resistor V25.25Ω Resistor V20.50Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E The V AB is the energy actaully supplied to the circuit AND what is applied across each parallel branch.

VIR Source Totals 9.00 V1.35 A6.65Ω r 0.27 V1.35 A0.20Ω Resistor V I 1 = V 1 /R 1 = A 15.00Ω Resistor V I 2 = V 2 /R 2 = A 25.25Ω Resistor V I 3 = V 3 /R 3 = A 20.50Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E Ohm’s Law determines the portion of the total current flowing down each parallel branch.

VIR Source Totals 9.00 V1.35 A6.65Ω r 0.27 V1.35 A0.20Ω Resistor V A15.00Ω Resistor V A25.25Ω Resistor V I 3 = I T - (I 1 + I 2 ) = A 20.50Ω Parallel Circuits r R1R1 R2R2 R3R3 B A E There is an alternate solution for I 3 as it is the remaining current.