IMG_20191022_100450.jpg - 150 IN1 SERIES de CIRCUITS VOLTAGE DIVISION IN A SERIES CIRCUIT III 151 For path 2 starting at point a in a clockwise

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Unformatted text preview: 150 IN1 SERIES de CIRCUITS VOLTAGE DIVISION IN A SERIES CIRCUIT III 151 For path 2, starting at point a in a clockwise direction, - V2 - 20 V = 0 and V2 = - 20 V Solutions : Applying Kirchhoff's voltage law in the clockwise direction start- The minus sign in the solution simply indicates that the actual pola ties are different from those assumed. ing at the negative terminal of the supply results in - E+ V3 + V 2 + V 1 = 0 E = VI + V2 + V3 (as expected) The next example demonstrates that you do not need to know why off and so that V2 = E - V 1 - V3 = 54 V - 18 V - 15V elements are inside a container when applying Kirchhoff's voltage la V2 = 21 V They could all be voltage sources or a mix of sources and resistors. and doesn't matter-simply pay strict attention to the polarities encountere V2 _ 21 V Try to find the unknown quantities in the next examples witho R 2 trouble. looking at the solutions. It will help define where you may be having 12 = 3A V1 18 V - = 60 40 V Example 5.1 1 emphasizes the fact that when you are applying Kirch C, RI = 3A hoff's voltage law, the polarities of the voltage rise or drop are the im 1 = 50 portant parameters, not the type of element involved. with R3 = V3 _ 15 V 13 3 A 50 X EXAMPLE 5.11 Using Kirchhoff's voltage law, determine the un EXAMPLE 5.14 Using Kirchhoff's voltage law and Fig, 5.12, verify known voltage for the circuit in Fig. 5.30. Ed. ( 5.1 ). + 30 V - Solution: Note that in this circuit, there are various polarities across Solution: Applying Kirchhoff's voltage law around the closed path: FIG. 5.30 the unknown elements since they can contain any mixture of compo. E = VI+ V2+ V3 Series configuration to be examined results in ments. Applying Kirchhoff's voltage law in the clockwise direction in Example 5.11. Substituting Ohm's law: +60 V - 40 V - Vx+ 30V=0 I, RT = 1R1 + 12 R2 + 13 R3 and V x = 60 V + 30 V-40 V =90V-40 V but Is = 1 1 = 12= 13 with V , = 50 V so that I, RT = 1, ( R1 + R2 + R3 ) and RT = R1+ R2 + R3 EXAMPLE 5.12 Determine the voltage Vx for the circuit in Fig. 5.31. which is Eq. (5.1). Note that the polarity of V, was not provided . + 2V - Solution: For cases where the polarity is not included, simply make an 5.7 VOLTAGE DIVISION IN A SERIES CIRCUIT FIG. 5.31 assumption about the polarity, and apply Kirchhoff's voltage law as be- Applying Kirchhoff's voltage law to a circuit in fore. If the result has a positive sign, the assumed polarity was correct. If The previous section demonstrated that the sum of the voltages which the polarities have not been provided for one the result has a minus sign, the magnitude is correct, but the assumed across the resistors of a series circuit will always equal the applied of the voltages ( Example 5.12). polarity must be reversed. In this case, if we assume point a to be posi- voltage. It cannot be more or less than that value. The next question is, How tive and point b to be negative, an application of Kirchhoff's voltage law will a resistor's value affect the voltage across the resistor? It turns out that n the clockwise direction results in the voltage across series resistive elements will divide as the magnitude V3 = 15 V -6 V -14 V - Vx+ 2V=0 of the resistance levels. and 1360 12 V so that V x = - 20 V + 2V In other words , 54V V x = - 18 V in a series resistive circuit, the larger the resistance, the more of the 2 70 V2 E =20 V 2 2 < 3 0 6V Since the result is negative, we know that point a should be negative and applied voltage it will capture. point b should be positive, but the magnitude of 18 V is correct. In addition , R, \$ 10 2V VI = 18 V the ratio of the voltages across series resistors will be the same as the EXAMPLE 5.13 For the series circuit in Fig. 5.32. ratio of their resistance levels. FIG. 5.32 Series configuration to be examined in . Determine V2 using Kirchhoff's voltage law. All of the above statements can best be described by a few simple ex- Example 5.13. b. Determine current 12. amples. In Fig. 5.33, all the voltages across the resistive elements are pro- vided. The largest resistor of 6 0 captures the bulk of the applied voltage, FIG. 5.33 c. Find R, and R3. while the smallest resistor, R3, has the least. In addition, note that since the Revealing how the voltage will divide across series Introductory, C.- 118 resistance level of R, is six times that of R3, the voltage across R, is six resistive elements....
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