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Engineering Science EAB_S_127 Electricity Chapter 2

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Circuits and Resistance The simplest circuit is one with an electric source (a cell here) and a resistor, which are connected with metal wires, as shown in Figure 2.1. As mentioned previously, the flow of electrical current through a resistance causes a voltage drop (as with the internal resistance of a cell) Figure 2.1 Electric circuit Metal wire + Cell V - RI

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Ohm’s Law In the early 19 th century Georg Ohm proved that the amount of current through a resistance is proportional to the voltage across it and inversely proportional to its resistance. Hence This can also be written as: Thus; if there is 1Volt across a resistor and it has 1A flowing through it, the resistor = 1 Ohm [ ] Example: 220V is supplied to a DC motor which has a resistance of 22 Ω. Find the current flowing in the motor? Answer: I = V/R = 220/22 = 10Amps

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+ A - I2I2 I3I3 I1I1 R1R1 R2R2 R3R3 ITIT VTVT Kirchoff’s Current Law Kirchoff’s current law states that, “The sum of all the currents passing through any node in a circuit equals zero”. This is the same as saying, “the sum of currents flowing into that node is equal to the sum of currents flowing out of that node”. See figure Hence at Node A;

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+ A - I2I2 I3I3 I1I1 R1R1 R2R2 R3R3 ITIT VTVT Kirchoff’s Current Law: Example 1 Given that I T = 6 Amps, I 1 = 2 Amps, I 2 = 1 Amp What is the value of I 3 ? Answer: I 3 = I T – I 1 – I 2 = 6 – 2 – 1 = 3 Amps

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+ A - I2I2 I3I3 I1I1 R1R1 R2R2 R3R3 ITIT VTVT Kirchoff’s Current Law: Example 2 Given that I 1 = 1.5 Amps, I 2 = 0.5 Amps, I 3 = 1 Amp What is the value of I T ? Answer: I T = I 1 + I 2 + I 3 = 1.5 + 0.5 + 1 = 3 Amps

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Kirchoff’s Voltage Law Kirchoff’s voltage law states that, “The sum of all the voltages around a closed loop equals zero”. This is the same as saying, “the sum of voltage drops in a loop is equal to the sum of voltage rises in a loop” V1V1 V2V2 + + + – – – V3V3

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Kirchoff’s Voltage Law: Example 1 Given that V 3 = 10 Volts and V 2 = 4 Volts What is the value of V 1 ? Answer: V 1 = V 3 – V 2 = 10 – 4 = 6 Volts V1V1 V2V2 + + + – – – V3V3

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Kirchoff’s Voltage Law: Example 2 Given that V 1 = 2 Volts and V 2 = 7 Volts What is the value of V 3 ? Answer: V 3 = V 1 + V 2 = 2 + 7 = 9 Volts V1V1 V2V2 + + + – – – V3V3

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Series Resistance When resistors are connected end to end they are said to be in series The total resistance can be shown to be the sum of the individual resistances I T = I 1 = I 2 = I 3 and V T = V 1 + V 2 + V 3 V T = I 1 R 1 + I 2 R 2 + I 3 R 3 = I T (R 1 + R 2 + R 3 ) Thus:- R T = V T /I T = R 1 + R 2 + R 3 VTVT ITIT V1V1 V2V2 V3V3 I1I1 I2I2 I3I3

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Series Resistance: Example 1 Given that, in the circuit below; R 1 = 10 , R 2 = 20 and R 3 = 30 What is the total series resistance? Answer: R T = 10 + 20 + 30 = 60 VTVT ITIT R1R1 R2R2 R3R3 I1I1 I2I2 I3I3

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Series Resistance: Example 2 Given that, in the circuit below; What is the total series resistance? What is the current I ? Answer: R T = 2 + 3 = 5 Answer: I = 5/5 = 1 Amp + 5 V5 V - I 2 Ω 3 Ω

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Resistors in Parallel When components are connected in the same way at both ends, they are said to be in parallel (e.g. resistors in figure below) Applying Kirchoff’s Laws give us I T = I 1 + I 2 + I 3 and V T = V 1 = V 2 = V 3 + VTVT - V1V1 I2I2 I3I3 I1I1 V2V2 V3V3 R1R1 R2R2 R3R3 ITIT

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Resistors in Parallel (2) Applying Ohm’s Law to each resistor gives: Since V T = V 1 = V 2 = V 3, they can be cancelled out and hence we have Alternatively this can be re-written specifically for three resistors as:

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Resistors in Parallel: Example 1 Given the circuit below, what is the parallel resistance? Answer: Alternatively:

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Resistors in Parallel: Example 2 Given the circuit below, what is the parallel resistance? Answer: Alternatively:

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Summary Learning Outcomes: Ohm’s Law Kirchoff’s Current Law Kirchoff’s Voltage Law Series Resistance Parallel Resistance Next Week: Wheatstone Bridge and Capacitance

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5 Parallel Circuits Chapter Topics Covered in Chapter 5

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