CH2_Problems - 1 08 C hapter 2 Resistive Circuits P roblems...

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108 Chapter 2 Resistive Circuits Problems Section 2.1: Resistances in Series and Parallel *P2.1. Reduce each of the networks shown in Figure P2.1 to a single equivalent resistance by com- bining resistances in series and parallel. 3Q 30Q 12Q 24Q 7Q 4Q (a) lOQ 6Q 1s n 60Q 9Q sn (b) Figure P2.1 *P2.2. A 4-Q resistance is in series with the paral- lel combination of a 20-Q resistance and an unknown resistance Rx. The equivalent resis- tance for the network 8 Q. Determine value of Rx. *P2.3. Find equivalent resistance looking into terminals a and bin Figure P2.3. Figure P2.3 *P2.4. Suppose that we need a resistance of 1.5 kQ and you have a box 1-kQ resistors. Devise a network of 1-kQ resistors so the equivalent resistance 1.5 kQ. Repeat for equivalent resistance of 2.2 kQ. *P2.5. equivalent resistance between ter- minals a Figure P2.5. Figure P2.5 P2.6. equivalent resistance between termi- nals a b for each of the networks shown in Figure P2.6. b 16!2 20!2 180 n (a) a b 3n (b) Figure P2.6 63!2 * Denotes answers are contained in the Student Solutions files. See Appendix F for more information about accessing the Solutions.
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4Q 30Q 24Q 6Q 20Q (c) Figure P2.6 (Cont.) P2.7. What resistance in parallel with 120 Q results an equivalent of 48 Q? P2.8. a. Determine the between termi- nals a and b for network shown in Figure P2.8. b. Repeat after connecting c d with a short circuit. ao------i ---------{) c bo-- ----- Figure P2.8 P2.9. Two resistances having values R 2R are parallel. R resis- tance both integers. possible values for R? P2.10. A connected terminals a b consists two parallel combinations that in series. Th e first parallel combi- nation is compo sed a 16-Q resistor a 48-Q resistor. The se cond parallel combina- tion composed a 12-Q resistor a 24-Q resistor. Draw determine its resistance. P2.11. Two resistances R1 R2 in parallel. We know R1 = 90 Q current through R2 is three times value R1. R2. P2.12. Find infi- nite shown Figure P2.12(a). Becau se its form, this is called a semi-infinite ladder. (Hint: If another Problems 1 09 section added to ladder as in Figure P2.12(b ), is same. Thus , working from Figure P2.12(b ), we can write expression for R eq terms Req. Then, solve Req.) JQ > lQ 2Q (a) (b) Figure P2.12 Ladder (a) • • • P2.13. connect n 1000-Q resistances parallel, what resistance? P2.14. heating element electric cook top has two resistive elements, R1 = 57.6 Q R2 = 115.2 Q, be operated separately, in series, or parallel from volt- ages either 120 V 240 V. For low- est power, is series with R 2, combination is from 120V. is lowest power? highest power, how should elements operated? power results? List more modes operation resulting for each. P2.15. We designing electric space heater operate from 120 V. Two heating elements with resistances R1 R2 used parallel, separately, in series. highest 1280 W, is 240 W. values needed R1 R 2? intermediate settings available? P2.16. Sometimes , we can use symmetry consid- erations find a circuit cannot reduced by series paral- lel combinations. A classic problem this type is illustrated Figure P2.16. Twelve
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110 Chapter 2 Resistive Circuits 1-Q resistors are arranged on the edges of a cube, and terminals a bare connected to diagonally opposite corners cube.
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CH2_Problems - 1 08 C hapter 2 Resistive Circuits P roblems...

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