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Engineering Science EAB_S_127 Electricity Chapter 2.

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1 Engineering Science EAB_S_127 Electricity Chapter 2

2 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

3 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

4 + 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;

5 + 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

6 + 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 = = 3 Amps

7 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 V2V – – – V3V3

8 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 V2V – – – V3V3

9 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 = = 9 Volts V1V1 V2V – – – V3V3

10 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

11 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 = = 60  VTVT ITIT R1R1 R2R2 R3R3 I1I1 I2I2 I3I3

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

13 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

14 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:

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

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

17 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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