Electric Circuits. Week. Simple Resistive Circuits )ا ششح األفىاس ا شئ ١ غ ١ ح(

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1 Electric Circuits Week Simple Resistive Circuits )ا ششح األفىاس ا شئ ١ غ ١ ح( خ ا غ ١ شو ١ د ذرى ا خ ػششج أعات ١ غ. ٠ حر و خ ػ ششح ح ي )51( عؤاال سواخ ا رحا اخ عاتمح. ض ا ا ج دج )03 ٠ ا ( ذ إػذاد زا ا خ و ذى شجؼا أعاع ١ ا رف له ا ثا ش تا ادج. إ ذحض سضان أػذ ا اعرؼذ ث ا. ذاخ األ ث ح ا ر اس ٠ اإلضاف ١ ح ٠ حر ا لغ )eng-hs.net( ػ فاخ إضاف ١ ح تى ا ح ي ؼشش ٠ عؤاال أخش سواخ ا رحا اخ عاتمح. ارا ال ذمر األحذز ا خ ٠ ر ذ م ١ ح ا ذحذ ٠ ث ا تشى أعث ػ ساجغ )eng-hs.net( رأوذ ششائه اإلصذاس األحذز. ا ؼج ١ ة أ أج ا ٠ حذز ا ال ؼثش ػ ١ ت رؼثش ت. اكسنلاتنملنفوةنبتسعونمدعنمنفينتلنمننلصنةوناكفوعنأمامناكهسدتس نأتسفلنصاكنننوسةمن نأننلصنةوناكجمعة ناكوئةتسة نباكتسودابنأتسفلنبةاسنن )1(

2 Resistors in Series: Sept 2015 Series-connected circuit elements carry the same current. Thus we can redraw Fig. 3.1 as shown in Fig Figure 3 Resistors connected in series. Figure 3.2 Series resistors with a single unknown current. To find, we apply Kirchhoff s voltage law around the loop: The seven resistors can be replaced by a single resistor. Thus we can redraw Fig. 3.2 as shown in Fig Figure 3. A simplified version of the circuit shown in Fig ا غضة ر ه ا غ ا ز ٠ ثر ؼ ا شء آ ال أ ٠ خ غ ١ ش. )2(

3 Resistors in Parallel: The defining characteristic of parallel-connected elements is that they have the same voltage across their terminals. From Kirchhoff's current law, The voltage across each resistor must be the same. Hence, from Ohm's law, Therefore, Figure 3.5 Resistors in parallel. Substituting Eq. 3.9 into Eq. 3.7 yields Figure 3.6 Nonparallel resistors. Note that the resistance of the equivalent resistor is always smaller than the resistance of the smallest resistor in the parallel connection. Many times only two resistors are connected in parallel. Figure 3.8 illustrates this special case. Or )3(

4 The Voltage-Divider and Current-Divider Circuits: Applying Kirchhoff's voltage law around the closed loop yields: Or Figure 3.12 (a) A voltage-divider circuit and (b) the voltage-divider circuit with current i indicated. The Current-Divider Circuits: The current-divider circuit shown in Fig consists of two resistors connected in parallel across a current source. Figure 3.15 The current-divider circuit. The current divider is designed to divide the current between and. We find the relationship between the current and the current in each resistor by directly applying Ohm's law and Kirchhoff s current law The voltage across the parallel resistors is: إ ا أ ذفؼ ا ش ء أ ال ذفؼ وف ذشددا. )4(

5 Voltage Division and Current Division: Sept 2015 We start by using Ohm's law to calculate through all of the resistors in series: We apply Ohm's law a second time to calculate the voltage drop across the resistor, using the current calculated: We start by using Ohm's law to calculate across each of the resistors in parallel: ( ) Figure 3.19 Circuit used to illustrate current division. Note that the constant of proportionality in the current division equation is the inverse of the constant of proportionality in the voltage division equation. جح ١ ا ذ ١ ا أ ٠ ى ا شء عج ١ خ ف أ ذشدد تؼض ا اط ٠ ج غ ت ١ ا. )5(

6 Measuring Voltage and Current: When working with actual circuits, you will often need to measure voltages and currents. We will spend some time discussing several measuring devices here and in the next section, because they are relatively simple to analyze and offer practical examples of the current- and voltage-divider configurations we have just studied. An ammeter is an instrument designed to measure current; it is placed in series with the circuit element whose current is being measured. A voltmeter is an instrument designed to measure voltage; it is placed in parallel with the element whose voltage is being measured. There are two broad categories of meters used to measure continuous voltages and currents: digital meters and analog meters. Digital meters measure the continuous voltage or current signal at discrete points in time, called the sampling times. Analog meters are based on the d'arsonval meter movement which implements the readout mechanism. For example, one commercially available meter movement is rated at 50 mv and 1 ma. This means that when the coil is carrying 1 ma, the voltage drop across the coil is (50 mv) and the pointer is deflected to its full scale position. A Fig 3.23 schematic illustration of a d'arsonval meter movement is shown in Fig An analog ammeter consists of a d'arsonval movement in parallel with a resistor, as shown in Fig The purpose of the parallel resistor is to limit the amount of current in the movement's coil by shunting some of it through. An analog voltmeter consists of a d'arsonval movement in series with a resistor, as shown in Fig Here, the resistor is used to limit the voltage drop across the meter's coil. In both meters, the added resistor determines the fullscale reading of the meter movement. Fig 3.2 Fig 3.2 )6( و ا صؼذ أحذ تاذجا ا م ح ذط ع وث ١ ش غحث إ ا ماع.

7 Measuring Resistance-the Wheatstone Bridge: The Wheatstone bridge circuit is used to precisely measure resistances of medium values, that is, in the range of ( Ω) to ( MΩ). In commercial models of the Wheatstone bridge, accuracies on the order of (±0.1%) are possible. The bridge circuit consists of four resistors, a dc voltage source, and a detector. The resistance of one of the four resistors can be varied, which is indicated in Fig by the arrow through. The dc voltage source is usually a battery, which is indicated by the battery symbol for the voltage source in Fig The detector is generally a d'arsonval movement in the micro amp range and is called a galvanometer. Figure 3.26 shows the circuit arrangement of the resistances, battery, and detector where and are known resistors and is the unknown resistor. Fig 3.2 Sept 2015 To find the value of, we adjust the available resistor untill there is no current in the galvanometer. We then calculate the unknown resistor from the simple expression When is zero, that is, when the bridge is balanced, Kirchhoff's current law requires that Now because is zero, there is no voltage drop across the detector, and therefore points (a) and (b) are at the same potential. Thus when the voltage is balanced, Kirchhoff's voltage law requires that Solving, From which, )7( ٠ ى ه ذحم ١ ك ش ء ذؼرمذ أ ه ػاجض ػ ذحم ١ م.

8 Solution: a) For circuit (a): For circuit (b): For circuit (c): b) ٠ غ ظ أغ ة ا اط ف ا رز ش ح ي أش ١ اء ال ٠ ى ا إ ٠ غ ا ىث ١ ش ا ز ٠ ى. )8(

9 Solution: a) b) ٠ حاسب ا جثا ى ر ه (c ح ١ ال ٠ ر ى ا شب. )9(

10 a) [ ] b) c) Thus the power developed by the current source is أ ح ١ ا ر ه ش ء ادس ا حذ ز ا حم ١ مح أ ؼظ ا اط ر اجذ فمظ زا و ش ء. )11(

11 Solution: ترطث ١ ك ا (KCL( ػ ذ و node حص ػ ل ١ ا ر ١ اس ف و ما و ا تا شى ا را. i i i a b c 5 ma i R i R i R i R.) فشق ا ج ذ ػ ج ١ غ ا ما اخ فغ ٠ غا ( ٠ ذ اإل غا جا ال ١ ظ غث ١ ا ى أغ ة أ ظ ح ا رؼ ١ ذح إ غث. اكسنلاتنملنفوةنبتسعونمدعنمنفينتلنمننلصنةوناكفوعنأمامناكهسدتس نأتسفلنصاكنننوسةمن نأننلصنةوناكجمعة ناكوئةتسة نباكتسودابنأتسفلنبةاسنن )11(

12 Solution: a) Getting equivalent resistance for Ω) and Ω) which are in parallel: Getting equivalent resistance for ( Ω) and ( Ω + Ω) which are in parallel: Getting equivalent resistance for ( Ω) with ( Ω + Ω) which are in parallel: Using current divider: (b ) ( ا لاحح أ ذغأي صذ ٠ مه ا ز لغ ف سطح إ وا ٠ حراج غاػذذه. )12(

13 Solution: a) Getting eqwoalent resistance for and winch are in parallel: ( ما ر ١ Ω 03 Ω, 1 تحغاب ) م تحغاب ا ر ١ اس ى ما ح ثاششج ػ ا شع و ا تا شى : 5 A 4.5 A 3A 0.5A 0.5A 2A 1A 5A 4A 1A Applying KVL for the left loop: )13( لدذ ٠ شده ا داط ف ١ دا ذمد ي ى ع ١ صذ ل ا ذفؼ.

14 Solution: We can change the above ( to Υ) as shown: R R R Ω 8 Ω Now we can get ( ) for all resistances: () ( ) ا ؼ ش ء سائغ إرا ذى ذمراخ ػ ١. )14(

15 Solution: As along as the voltage on the detector careless can be considered as ashort circuit Using current (a, b): Appling ( KCL) at middle node: ٠ خ أحذ تذال ه ف رح ١١ ا ح ١ اذه تا طش ٠ مح ا ر ذشا ا أ د اعثح ال و ا ٠ شا ا. )15(

16 Delta-to-Wye (Pi-to-Tee) Equivalent Circuits: The resistors,, and in the circuit shown in Fig are referred to as a delta ( ) interconnection because the interconnection looks like the Greek letter ( ). It also is referred to as a pi interconnection Figure 3.28 A configuration Sept 2015 configuration viewed as a because the ( ) can be shaped into a π without disturbing the electrical equivalence of the two configurations. The resistors,, and in the circuit shown in Fig are referred to as a wye (Y) interconnection because the interconnection can be shaped to look like the letter (Y). The (Y) configuration Figure 3.30 A Y structure viewed as a T structure. also is referred to as a tee (T) interconnection because the (Y) structure can be shaped into a (T) structure without disturbing the electrical equivalence of the two structures. Figure 3.31 illustrates the ( -to-y) (or π -to-t) equivalent circuit transformation. Note that we cannot transform the ( ) interconnection into the (Y) interconnection simply by changing the shape of the interconnections. Figure 3.31 A -to-y transformation. )16( إرا مذن ا اط ف ؼ ا أ ه فؼ د ا ٠ غرحك ا زوش.

17 Saying the -connected circuit is equivalent to the Y-connected circuit means that the configuration can be replaced with a Y configuration to make the terminal behavior of the two configurations identical. The Y-connected resistors in terms of the -connected resistors: The three -connected resistors as functions of the three Y connected resistors are: أغ ة ا اط ال ٠ د أ ٠ صثح ا أغ ١ اء فحغة ت أ ٠ صثح ا أغ ا ٢ خش ٠. )17(

18 We can (Υ-to- ) transformation to reach the following shape: We can get the equivalent of each two parallel resistance: دع تؼض أخطاء ا ٢ خش ٠ ذ ش د أ ذالحظ ا. )18(

19 Continued (Solution problem 3.54): a) b) c) Applying (KCL) at the right nook in original circuit: Applying (KVL) for the right loop: We can get the equivalent resistance for the whole circuit: We can get the power delivered: Thus, the current source delivers W. وأ ا حاعذ إ ا خ ك ١ غراظ. )19(

20 Solution: a) ( ) ال ذ لف طف ه ػ ذ حذ د ذؼ ه ف د ف ص خر ف. )21(

21 Continued Solution(problem 3.56): ) [[ ] ] [ ] c) Convert the delta connection to it' sequivalent wye. Convert the wye connection ( ) to it' sequivalent delta. ال ٠ غرط ١ غ أحذ سو ب ظ شن إال إرا و د ح ١ ا. )21(

22 Solution: Converting the left,, to Y ) : + - ( ) c) Using current divider: d) a) Applying (KVL) for the upper right loop: b) ا غ ء جذا أ ٠ غر ج دن ػذ ػ ذ تؼض ا مشت ١ ه. )22(