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DET 101/3 Basic Electrical Circuit 1 DC CIRCUITS: CHAPTER 2 RESISTIVE CIRCUITS Series/Parallel Equivalent Circuits Voltage Divider Rule (VDR) Current Divider Rule (CDR) Voltage and Current Measurements Wheatstone Bridge Delta (or Pi) and Wye (or Tee) Equivalent Circuit Series/Parallel Equivalent Circuits Most common connection found in circuit analysis. Circuit simplifying technique. Several resistors are combined to represent a single equivalent resistance. Equivalent resistance depends on two (2) factors: Type of connection Point of terminals Series Equivalent Circuit The equivalent resistance for any number of resistors in series connection is the sum of each individual resistor. x x R1 R2 R3 R4 RN y Req R5 y N Req R1 R2 R3 R4 R5 RN Rn n 1 (2.1) Series Equivalent Circuit (Continued…) Apparently the single equivalent resistor is always larger than the largest resistor in the series connection. Series resistors carry the same current thru them. Voltage across each of the resistors obtained using voltage divider rule principle or Ohm’s law. Parallel Equivalent Circuit The equivalent resistance for any number of resistors in parallel connection is obtained by taking the reciprocal of the sum of the reciprocal of each single resistor in the circuit. x x R1 R2 R3 R4 R5 RN Req y y 1 1 1 1 1 Req RN R1 R2 R3 R4 R5 1 1 n 1 Rn N 1 (2.2) Parallel Equivalent Circuit (Continued…) Apparently, the single equivalent resistor is always smaller than the smallest resistor in the parallel connection. Voltage across each resistor must be the same. Currents thru each of them are divided according to the current divider rule principle. Parallel Equivalent Circuit (Continued…) Special simplified formula if the number of resistors connected in series is limited to two elements i.e. N=2. Req R1 R2 R1 R2 (2.3) **When just two resistors connected in parallel the equivalent resistance is simply the product of resistances divided by its sum. Special Cases of Connections: Open Circuit (O.C) An opening exists somewhere in the circuit. The elements are not connected in a closed path. i=0A R1 R2 Vs R3 a b O.C: i = 0 A KVL: Voc = Vs Ohm’s Law: Rab = V/I = ∞ Special Cases of Connections: Short Circuit (S.C) Both of its terminal are joint at one Vs single node. The element is bypassed. i a R1 i 0A R2 R3 b S.C: Rab = 0 Ohm’s Law: i = Vs/(R1 + R3) : Vsc = 0 V Practice Problem 2.9 Q: By combining the resistors in Figure below, find Req. Practice Problem 2.10 Q: Find Rab for the circuit in Figure 2.39. Practice Problem 2.11 Q: Calculate Geq in the circuit of Figure 2.41. Voltage Divider Rule (VDR) Whenever voltage has to be divided among resistors in series use voltage divider rule principle. + V1 R1 R2 R3 + V3 - + V2- Vs R1 V1 Vs R1 R2 R2 V2 Vs R1 R2 VDR (Continued…) In general, to find the voltage drop across the nth resistor in the voltage divider circuit configuration we use this formula: Rn Vn Vs R1 R2 R3 RN (2.4) Where n = 1, 2, 3,.....N Practice Problem 2.12 Find V1 and V2 in the circuit shown in Figure 2.43. Also calculate i1 and i2 and the power dissipated in the 12 and 40 resistors. Current Divider Rule (CDR) Whenever current has to be divided among resistors in parallel, use current divider rule principle. is i1 Vs R1 R2 i1 is R1 R2 i2 R2 R1 i2 is R1 R2 CDR (Continued…) Circuit with more than two branches… is Vs i1 i2 i3 R1 R2 R2 G1 i1 is G1 G2 G2 i2 is G1 G2 • In general, for N-conductors the formula represents: Gn in is G1 G2 G3 GN (2.5) n = 1, 2, 3…..N Practice Problem 2.13 Find (a)V1 and V2 (b) the power dissipated in the 3 k and 20 k resistors and (c) power supplied by the current source. 1 k 3 k + V1 - 10 mA 5 k Figure 2.45 + V2 - 20 k Chapter 2, Problem 34 Determine i1, i2, v1, and v2 in the ladder network in Fig. 2.98. Calculate the power dissipated in the 2- resistor. Chapter 2, Problem 36 Calculate Vo and Io in the circuit of Fig. 2.100. Voltage and Current Measurements To determine the actual and quantitative behavior of the physical system. Two most frequently used measuring devices in the laboratories: Ammeter Voltmeter Ammeter Must be placed in series connection with the element whose current is to be measured. An ideal ammeter should have an equivalent resistance of 0 and considered as short circuit equivalent to the circuit where it is being inserted. Voltmeter Must be placed in parallel connection with the elements whose voltage is to be measured. An ideal voltmeter should have an equivalent resistance of ∞ and considered as open circuit equivalent to the circuit where it is being inserted. Meter Types Analog meters Based on the d’Arsonval meter movements. Digital meters More popular than analog meters. More precision in measurement, less resistance and can avoid severe reading errors. Measure the continuous voltage or current at discrete instants of time called sampling times. Configuration of Voltmeter and Ammeter In A Circuit RA = 0 R1 Vs A R2 V RV = Inf Wheatstone Bridge : Practical Application of Resistance Measurement Invented by a British professor, Charles Wheatstone in 1847. More accurate device to measure resistance in the mid-range (1 to 1 M) In commercial models of the Wheatstone bridge, accuracies about ± 0.1% are achievable Wheatstone Bridge (Continued…) The bridge circuit consists of R1 R2 G a R3 Ig b x V + V3 Figure A - + Four resistors A dc voltage source A detector known asVs galvanometer (microampere range) Rx Balanced Bridge If R3 is adjusted until the current Ig in the galvanometer is zero the bridge its balance state. No voltage drop across the detector which means point a and b are at the same potential. Implies that V3 = Vx when Ig = 0 A. Balanced Bridge (Continued…) Applying the voltage divider rule (VDR): R3 V3 Vs R1 R3 and Rx Vx Vs R2 Rx Balanced Bridge (Continued…) Since no current flows through the galvanometer, V3 Vx R3 Rx Vs Vs R1 R3 R2 Rx R2 R3 Rx R1 hence R2 Rx R3 R1 (2.6) Example 1 The galvanometer shows a zero current through it when Rx measured as 5 k. What do you expect to be the value of the adjustable resistor, R3? Show your derivation in getting the formula. 2 k Ig G V3 V - + R3 b x a + Vs 2k5 Rx Exercise 1 The bridge in Figure A is energized by 6V dc source and balanced when R1 = 200, R2 = 500 and R3 = 800. (a) What is the value of Rx? (b) How much current (in miliamperes) does the dc source supply? (c) Which resistor absorbs the least power and which absorbs the most? How much? Unbalanced Bridge To find Ig when the Wheatstone bridge is unbalanced, use Thevenin equivalent circuit concept to the galvanometer terminals. Assuming Rm is the resistance of the galvanometer yields, Ig Vth Rth Rm (2.7) Delta (or Pi) and Wye (or Tee) Equivalent Circuit Stuck with neither series nor parallel connection of the resistors in part of a circuit. Simplify the resistive circuit to a single equivalent resistor by means of threeterminal equivalent circuit. Wye/Tee Circuit Same type of connections R1 a b R1 R2 a R2 b R3 R3 c c (a) Wye c (b) Tee Delta/Pi Circuit Same type of Connections Rc Rc a b Rb a Ra Rb Ra c c (a) b Delta c (b) Pi Delta-to-Wye and Wye-to-Delta Transformation Remember that before and after transformation using either Wye-to-Delta or Delta-to-Wye, the terminal behavior of the two configurations must retain. Then only we can say that they are equivalent to each other. Special Case of -Y Transformation A special case occur when R1 = R2 = R3 = RY or Ra = Rb = Rc =R under which the both networks are said to be balanced. Hence the transformation formulas will become: RY = R/3 or R = 3RY By applying Delta/Wye transformations, we may find that this final process leads to series/parallel connections in some parts of the circuit. Delta to Wye Transform To obtain the equivalent resistances in the Wye-connected circuit, we compare the equivalent resistance for each pair of terminals for both circuit configurations. Rc ( Ra Rb ) connected : Rab Rc || ( Ra Rb ) Ra Rb Rc Y connected : Rab R1 R2 Delta to Wye Transform(Continued…) To retain the terminal behavior of both configurations i.e. R = RY So that, Rc ( Ra Rb ) Rab R1 R2 Ra Rb Rc (2.9) (2.8) Ra ( Rb Rc ) Rbc R2 R3 Ra Rb Rc Rb ( Ra Rc ) Rca R1 R3 Ra Rb Rc (2.10) Delta to Wye Transform(Continued…) To obtain the resistance values for Y-connected elements, by straightforward algebraic manipulation and comparisons of the previous three equations gives, Rb Rc R1 Ra Rb Rc (2.12) Ra Rb R3 Ra Rb Rc (2.11) Rc Ra R2 Ra Rb Rc (2.13) Wye to Delta Transform By algebraic manipulation, obtain the sum of all possible products of the three Y-connected elements; R1, R2 and R3 in terms of -connected elements; Ra, Rb and Rc.(From Eq. (2.11 – 2.13) Ra Rb Rc ( Ra Rb Rc ) R1 R2 R2 R3 R3 R1 2 ( Ra Rb Rc ) Ra Rb Rc ( Ra Rb Rc ) (2.14) Wye to Delta Transform(Continued…) Then we divide Eq. (2.14) by each of Eq. (2.11) to (2.13) to obtain each of the -connected elements as to be found variable in your left-side and its equivalent in Yconnected elements. (2.14) / (2.11) : Ra Rb Rc ( Ra Rb Rc ) R1 R2 R2 R3 R3 R1 Rb Rc R1 ( R R R ) b c a Wye to Delta Transform(Continued…) (2.14) / (2.11) : R1 R2 R2 R3 R3 R1 Ra R1 (2.15) Using the same manner, (2.14) / (2.12) : R1 R2 R2 R3 R3 R1 Rb R2 (2.16) (2.14) / (2.13) : R1 R2 R2 R3 R3 R1 Rc R3 (2.17) Wye-Delta Transformations Delta -> Star Star -> Delta Rb Rc R1 ( Ra Rb Rc ) Ra R1 R2 R2 R3 R3 R1 R1 Rc Ra R2 ( Ra Rb Rc ) Rb R1 R2 R2 R3 R3 R1 R2 Ra Rb R3 ( Ra Rb Rc ) Rc R1 R2 R2 R3 R3 R1 R3 Superposition of Delta and Wye Resistors “Each resistor in the Yconnected circuit is the product of the two resistors in two adjacent branches divided by the sum of the three a resistors” “Each resistor in the connected circuit is the sum of all possible products of Y resistors taken two at a time divided by the opposite Y resistors” Rc b Rb R1 R2 Ra R3 c Practice Problem 2.15 Q: For the bridge circuit in Fig. 2.54, find Rab and i. i a 13 10 100 V b Figure 2.54 50 Exercise 2 Use -to-Y transformation to find the voltages v1 and v2. 15 1 24 V + V1 - 10 + V2 - Exercise 3 Find the equivalent resistance Rab in the circuit below. Exercise 4 Find Rab in the circuit below. 9 k 9 k 9 k 9 k 9 k 9 k 9 k a b 9 k 9 k