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G.M . JAHEDUL ISLAM

Basic Electricity (6711). G.M . JAHEDUL ISLAM. Instructor Department of ELECTRICAL IDEAL INSTITUTE OF SCIENCE AND TECHNOLOGY (IIST). Previous Lesson. In my first class I discuss about concepts of electrical circuit. Concepts of electrical circuit. Definition of electrical circuit.

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G.M . JAHEDUL ISLAM

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  1. Basic Electricity (6711) G.M. JAHEDUL ISLAM Instructor Department of ELECTRICAL IDEAL INSTITUTE OF SCIENCE AND TECHNOLOGY (IIST)

  2. Previous Lesson In my firstclass I discuss about concepts of electrical circuit Concepts of electrical circuit. Definition of electrical circuit. Different types of electric circuit.

  3. Learning outcomes • After finished the lesson. students will be able know: • Series and parallel circuit. • Solutions to the problems relating to series and parallel circuits.

  4. Series circuits: • Series circuits • In a television series, you get several episodes, one after the other. A series circuit is similar. You get several components one after the other. • If you follow the circuit diagram from one side of the cell to the other, you should pass through all the different components, one after the other, without any branches.

  5. Series circuits • If you put more lamps into a series circuit, the lamps will be dimmer than before. • In a series circuit, if a lamp breaks or a component is disconnected, the circuit is broken and all the components stop working. Series circuits are useful if you want a warning that one of the components in the circuit has failed. They also use less wiring than parallel circuits.

  6. Series circuits The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit. Req = R1 + R2 + R3 = 17 Ω + 12 Ω + 11 Ω = 40 Ω Now that the equivalent resistance is known, the current at the battery can be determined using the Ohm's law equation. In using the Ohm's law equation (ΔV = I • R) to determine the current in the circuit, it is important to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here: Itot = ΔVbattery / Req = (60 V) / (40 Ω) = 1.5 amp The 1.5 amp value for current is the current at the battery location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the same as the current at each resistor location. Subsequently, the 1.5 amp is the value of I1, I2, and I3.

  7. Series circuits • Ibattery = I1 = I2 = I3 = 1.5 amp There are three values left to be determined - the voltage drops across each of the individual resistors. Ohm's law is used once more to determine the voltage drops for each resistor - it is simply the product of the current at each resistor (calculated above as 1.5 amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below. • ΔV1 = I1 • R1 ΔV1 = (1.5 A) • (17 Ω) • ΔV1 = 25.5 V • ΔV2 = I2 • R2 ΔV2 = (1.5 A) • (12 Ω) • ΔV2 = 18 V • ΔV3 = I3 • R3 ΔV3 = (1.5 A) • (11 Ω) • ΔV3 = 16.5 V • As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the voltage rating of the battery. In other words, is ΔVbattery = ΔV1 + ΔV2 + ΔV3 ? • Is ΔVbattery = ΔV1 + ΔV2 + ΔV3 ? Is 60 V = 25.5 V + 18 V + 16.5 V ? • Is 60 V = 60 V? • Yes!!

  8. Parallel circuits • Parallel circuits • In parallel circuits different components are connected on different branches of the wire. If you follow the circuit diagram from one side of the cell to the other, you can only pass through all the different components if you follow all the branches.

  9. Parallel circuits • In a parallel circuit, if a lamp breaks or a component is disconnected from one parallel wire, the components on different branches keep working. And, unlike a series circuit, the lamps stay bright if you add more lamps in parallel.

  10. Parallel circuits • Parallel circuits are useful if you want everything to work, even if one component has failed. This is why our homes are wired up with parallel circuits.

  11. Parallel circuits • Three resistors are connected in parallel. If placed in a circuit with a 12-volt power supply. Determine the equivalent resistance, the total circuit current, and the voltage drop across and current in each resistor

  12. Parallel circuits

  13. Parallel circuits • The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit. • 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 = (1 / 11 Ω) + (1 / 7 Ω ) + (1 / 20 Ω) 1 / Req = 0.283766 Ω-1 • Req = 1 / (0.283766 Ω-1) • Req = 3.52 Ω • (rounded from 3.524027 Ω) • Now that the equivalent resistance is known, the current in the battery can be determined using the Ohm's law equation. In using the Ohm's law equation (ΔV = I • R) to determine the current in the circuit, it is important to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here: • Itot = ΔVbattery / Req = (12 V) / (3.524027 Ω) Itot = 3.41 Amp • (rounded from 3.4051948 Amp) • The 12 V battery voltage represents the gain in electric potential by a charge as it passes through the battery. The charge loses this same amount of electric potential for any given pass through the external circuit. That is, the voltage drop across each one of the three resistors is the same as the voltage gained in the battery:

  14. Parallel circuits • ΔV battery = ΔV 1 = ΔV 2 = ΔV 3 = 12 V There are three values left to be determined - the current in each of the individual resistors. Ohm's law is used once more to determine the current values for each resistor - it is simply the voltage drop across each resistor (12 Volts) divided by the resistance of each resistor (given in the problem statement). The calculations are shown below. • I1 = ΔV1 / R1 I1 = (12 V) / (11 Ω) • I1 = 1.091 Amp • I2 = ΔV 2 / R2 I2 = (12 V) / (7 Ω) • I2 = 1.714 Amp • I3 = ΔV 3 / R3 I3 = (12 V) / (20 Ω) • I3 = 0.600 Amp • As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the current values for each individual resistor is equal to the total current in the circuit (or in the battery). In other words, is Itot = I1 + I2 + I3 ? • Is Itot = I1 + I2 + I3 ? Is 3.405 Amp = 1.091 Amp + 1.714 Amp + 0.600 Amp ? • Is 3.405 Amp = 3.405 Amp? • Yes!!

  15. The main concepts associated with series and parallel circuits are organized in the table below • Each of the above concepts has a mathematical expression. Combining the mathematical expressions of the above concepts with the Ohm's law equation (ΔV = I • R) allows one to conduct a complete analysis of a combination circuit.

  16. Feed back: • Distinguish series and parallel circuits. • Solutions to the problems relating to series and parallel circuits.

  17. Home work

  18. Next lecture: • Combination Circuits.

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