By Richard C Dorf; James A Svoboda

ISBN-10: 0471730424

ISBN-13: 9780471730422

**Read or Download Introduction to Electric Circuits [SOLUTIONS MANUAL] PDF**

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**Additional resources for Introduction to Electric Circuits [SOLUTIONS MANUAL]**

**Example text**

With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate 24 mW, 16 mW and 8 mW respectively. Therefore the design is complete. 3-26 DP3-2 Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage 200 division, 12 = 4 ⇒ R 2 = 400 Ω . 16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the is 400 8 series combination of two 200 Ω resistors. 08 W < W . 85% Hence, vo = 4 V ± 8% and the design is complete.

Using voltage 200 division, 12 = 4 ⇒ R 2 = 400 Ω . 16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the is 400 8 series combination of two 200 Ω resistors. 08 W < W . 85% Hence, vo = 4 V ± 8% and the design is complete. 6W ∴P= 10 v= DP3-4 N 1 N 1 = N i = G v = v where G = ∑ T T R R n = 1 Rn ∴N= iR ( 9 )(12 ) = = 18 bulbs v 6 3-27 28 Chapter 4 – Methods of Analysis of Resistive Circuits Exercises Ex. 3-1 v −v a a b + + 3 = 0 ⇒ 5 v − 3 v = −18 a b 3 2 v KCL at a: KCL at b: v −v b a − 3 −1 = 0 ⇒ v − v = 8 b a 2 Solving these equations gives: va = 3 V and vb = 11 V Ex.

25 = 1125 . 7-13 2 R( R ) 2 = R 2R + R 3 v 2 240 Pdeliv. 67 ∴Reported value was correct. 4 V o 320 + 650 + 230 ∴Reported value was incorrect. 2iH = 0 KCL at left node: iA + iB = iH This alone shows the reported results were incorrect. 49 A ∴ Reported values were incorrect. 5) = 14 V The KVL equations are satisfied so the analysis is correct. 5 mA That's a contradiction. The given values of ia and ib are not correct. Design Problems DP3-1 Using voltage division: vm = R 2 + aR p R1 + (1 − a ) R p + R 2 + aR p 24 = R 2 + aR p R1 + R 2 + R p 24 vm = 8 V when a = 0 ⇒ R2 R1 + R 2 + R p vm = 12 V when a = 1 ⇒ R2 + R p R1 + R 2 + R p The specification on the power of the voltage source indicates 242 1 ≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω R1 + R 2 + R p 2 = 1 3 = 1 2 Try Rp = 2000 Ω.

### Introduction to Electric Circuits [SOLUTIONS MANUAL] by Richard C Dorf; James A Svoboda

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