1 / 30

EEE205

EEE205. Yrd. Doç. Dr. Mehmet Ali Aktaş m ehmet.aktas@toros.edu.tr. Ohms Law.

leon
Télécharger la présentation

EEE205

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. EEE205 Yrd. Doç. Dr. Mehmet Ali Aktaş mehmet.aktas@toros.edu.tr

  2. OhmsLaw • Ohm’s law states that the current I flowing in a circuitis directly proportional to the applied voltage V andinversely proportional to the resistance R, provided thetemperatureremainsconstant. Thus, • Problem 1:The current flowing through a resistoris 0.8A when a p.d. of 20V is applied. Determinethe value of the resistance. FromOhm’slaw, resistance

  3. Problem 2. A 100V battery is connected acrossa resistor and causes a current of 5mA to flow.Determine the resistance of the resistor. If thevoltage is now reduced to 25V, what will be thenew value of the current flowing? Resistance Current when voltage is reduced to 25V,

  4. Series AndParallel Networks • Figure 5.1 shows three resistors R1, R2 and R3 connectedend to end, i.e. in series, with a battery sourceof V volts. Since the circuit is closed a current I willflowand the p.d. across each resistor may be determinedfrom the voltmeter readings V1, V2 and V3.

  5. Series AndParallel Networks • In a seriescircuit • The current I is the same in all parts of the circuitand hence the same reading is found on each oftheammetersshown, and • The sum of the voltages V1, V2 and V3 is equal tothe total applied voltage, V, i.e. V =V1+V2+V3 • From Ohm’s law: V1=IR1, V2=IR2, V3=IR3 andV =IR where R is the total circuit resistance. SinceV = V1 +V2 +V3 thenIR = IR1 + IR2 + IR3. DividingthroughoutbyI gives R=R1+R2+R3

  6. Series AndParallel Networks • Problem 1. For the circuit shown in Figure,determine (a) the battery voltage V , (b) the totalresistance of the circuit, and (c) the values ofresistors R1, R2 and R3, given that the p.d.’sacrossR1, R2 and R3 are 5V, 2V and 6V respectively.

  7. Series AndParallel Networks • Batteryvoltage • Total circuit resistance • Resistance Resistance Resistance

  8. Series AndParallel Networks • Problem 2. For the circuit shown in Fig. 5.3,determine the p.d. across resistor R3. If the totalresistance of the circuit is 100, determine thecurrent flowing through resistor R3. Find also thevalueof resistorR2 .

  9. Series AndParallel Networks P.d. acrossR3, V3=25−10−4=11V Current which is the current flowing in each resistor Resistance

  10. Series AndParallel Networks • Potentialdivider • The voltage distribution for the circuit shown inFig. (a) is given by:

  11. Series AndParallel Networks • The circuit shown in Fig.(b) is often referred to as apotential divider circuit. Such a circuit can consist ofa number of similar elements in series connected acrossa voltage source, voltages being taken from connectionsbetween the elements. Frequently the divider consists oftwo resistors as shown in Fig.(b), where:

  12. Series AndParallel Networks • Problem 4. Determine the value of voltage Vshownin Figure. • Figure may be redrawn as shown in followingFigure. • voltage

  13. Series AndParallel Networks ParallelNetworks • Figure shows three resistors, R1, R2 and R3 connectedacross each other, i.e. in parallel, across a batterysourceof V volts. • (a) the sum of the currents I1, I2 and I3 is equal to the total circuitcurrent, I , i.e. I = I1 +I2 +I3 and • (b) the source p.d., V volts, is the same across each oftheresistors. • FromOhm’slaw: • where R is the total circuit resistance. Since

  14. Series AndParallel Networks • Dividing throughout by V gives: • This equation must be used when finding the total resistanceR of a parallel circuit. For the special case of tworesistorsin parallel • Hence

  15. Series AndParallel Networks • For the circuit shown in Figure,determine (a) the reading on the ammeter, (b) the value of resistor R2. • P.d. across is the same as the supply voltage V.Hencesupplyvoltage, V =8×5=40V • (a) • (b) Current flowing through R2=11−8−2=1A.Hence

  16. Series AndParallel Networks • Problem . Two resistors, of resistance 3and6, are connected in parallel across a batteryhaving a voltage of 12V. Determine (a) the totalcircuit resistance and (b) the current in the 3resistor. The circuit diagram is shown in Figure • The total circuit resistance R is given by • Since than R=2 • Current in the 3resistance,

  17. Series AndParallel Networks • Given four 1resistors, state howthey must be connected to give an overall resistance of • (a) (b) 1(c) (d) • allfourresistors being connected in each case.

  18. Series AndParallel Networks • (a) All four in parallel (see Figure), since

  19. Series AndParallel Networks • (b) Two in series, in parallel with another two inseries (see Figure), since 1and 1in seriesgives 2, and 2in parallel with 2gives

  20. Series AndParallel Networks • (c) Three in parallel, in series with one (seeFigure), since for the three in parallel,

  21. Series AndParallel Networks • (d) Two in parallel, in series with two in series (seeFigure), since for the two in parallel.

  22. Series AndParallel Networks

  23. Series AndParallel Networks • Current division • For the circuit shown in Figure, the total circuitresistance, RT is given by:

  24. Series AndParallel Networks • From formula, ifwesimplifytheequationthanwe can calculatethewhich is flowingthroughand

  25. Series AndParallel Networks • Problem. Fortheseries-parallelarrangementshown in Figure, find • the supply current, • the current flowing through each resistor • the p.d. across each resistor.

  26. Series AndParallel Networks • (a) The equivalent resistance of and inparallelis: Ω • The equivalent resistance RT of R1, Rx and R4 inseriesis: • Supplycurrent

  27. Series AndParallel Networks • (b) The current flowing through and is 25A.The current flowing through The current flowing through

  28. Series AndParallel Networks • The equivalent circuit of Figure is: p.d. across, i.e. p.d. across, i.e. p.d. across, i.e.

  29. CapacitorsandCapacitance • A capacitor is an electrical device that is used to storeelectricalenergy. • Next to the resistor, the capacitor isthe most commonly encountered component in electricalcircuits. • Capacitors are used extensively in electricalandelectroniccircuits. • For example, capacitors are usedto smooth rectified a.c. outputs, they are used in elecommunicationequipment – such as radio receivers – fortuning to the required frequency, they are used in timedelay circuits, in electrical filters, in oscillator circuits,and in magnetic resonance imaging (MRI) in medicalbody scanners, to name but a few practical applications.

  30. CapacitorsandCapacitance

More Related