IMG_20191022_100233.jpg - SERIES 140 IN SERIES de CIRCUITS which for Fig 5.12 results in 109 10 0 VI = 1,Ri = 1,R =(60 mA(10 0 0.6 V A Vy lykg 1,ky(60

# IMG_20191022_100233.jpg - SERIES 140 IN SERIES de CIRCUITS...

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Unformatted text preview: SERIES 140 IN SERIES de CIRCUITS which for Fig. 5.12 results in 109 10 0 VI = 1,Ri = 1,R, = (60 mA )(10 0) - 0.6 V A Vy " lykg - 1,ky - (60 mA )(30 0) = 18 V RT R, \$ 100 0 140 6 V, " lyk, = 1,Ry = (60 mA )(100 0) - 6.0 V Note that in all the numerical calculations appearing in the text thus far, a unit of measurement has been applied to each calculated quantity. Always remember that a quantity without a unit of measurement is often FIG. 5.13 meaningless. Resistance "seen " at the terminals of a series circuit. EXAMPLE 5.4 For the series circuit in Fig. 5.15. R, -20 resistance "seen" at the connection terminals, as shown in Fig. 5.13(x In other words, it reduces the entire configuration to one such as in Fig a Find the total resistance Ry . Calculate the resulting source current ly 20 V 5. 13(b) to which Ohm's law can easily be applied. For the configuration in Fig. 5.12, with the total resistance calculated c. Determine the voltage across each resistor in the last section, the resulting current is E 8.4 V Solutions: 140 0 = 0.06 A = 60 mA -20+10+sn FIG. 5.15 Series circuit to be investigated in Example 5.4. Note that the current I, at every point or corner of the network is the RT - 80 same. Furthermore, note that the current is also indicated on the current display of the power supply. b. I m E 20 V RT 802 25 A Now that we have the current level, we can calculate the voltage C. V1 = 1,R, - 1,R1 - (25 A)(2 0) - 5V across each resistor. First recognize that Va = 12R2 - 1,Rx - (25 A)(1 0) = 25V the polarity of the voltage across a resistor is determined by the direction of the current. VJ = 13R3 - 1,R, - (25 AX(5 1) - 125V Current entering a resistor creates a drop in voltage with the polarity indicated in Fig. 5.14(a). Reverse the direction of the current, and the EXAMPLE 5.5 For the series circuit in Fig. 5.16: polarity will reverse as shown in Fig. 5.14(b). Change the orientation of a. Find the total resistance RT. the resistor, and the same rules apply as shown in Fig. 5.14(c). Applying b. Determine the source current /, and indicate its direction on the circuit. SO V the above to the circuit in Fig. 5.12 will result in the polarities appearing in that figure. c. Find the voltage across resistor Ry and indicate its polarity on the circuit. Solutions: 3. The elements of the circuit are rearranged as shown in Fig. 5.17. FIG. 5.16 Series circuit to be analyzed in Example S.5. v Z 10 0 RT = R2 + NR 7- 10n = 40 + (3)(70) (a) 10n * = 40 + 210 ( b ) RT = 25 0 FIG. 5.14 Inserting the polarities across a resistor as determined by the direction V, + of the current. wr The magnitude of the voltage drop across each resistor can then be 40 in That is, found by applying Ohm's law using only the resistance of each resistor. E = 50 V = RT VI = IRI V1 = 12R2 VJ = 13R3 (5.4) FIG, 5.17 Circuit in Fig. 5.16 redrawn to permit the use of Eq. (5.2)....
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• Fall '19
• SK Abid

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