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## Dorf homework solutions

All the material for this course will be covered during lectures completely. I Circuit v – R R v 2 kΩ 4 kΩ 6V 2V Fure P 5.4-8 Solution: From the given data: 2000 ⎫ voc ⎪ R t 2000 ⎪ ⎧ voc = 1.2 V ⎬ ⇒ ⎨ 4000 ⎩ R t = −1600 Ω voc ⎪ 2= R t 4000 ⎪ ⎭ 6= When R = 8000 Ω, v= R voc Rt R v= 8000 (1.2 ) = 1.5 V −1600 8000 P 5.4-13 The circuit shown in Fure P 5.4-13 consists of two parts, the source (to the left of the terminals) and the load.

6Ω – 10 Ω a 3Ω 18 V b Fure P 5.4-4 Find Rt: Rt = Write mesh equations to find voc: 12 (10 2 ) =6Ω 12 (10 2 ) Mesh equations: 12 i1 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i1 = 18 36 i1 = 18 ⇒ i1 = i2 = Finally, 1 A 2 14 ⎛ 1 ⎞ 7 ⎜ ⎟= A 3 ⎝2⎠ 3 ⎛7⎞ ⎛1⎞ voc = 3 i 2 10 i1 = 3 ⎜ ⎟ 10 ⎜ ⎟ = 12 V ⎝ 3⎠ ⎝ 2⎠ (checked using LNAP 8/15/02) P 5.4-8 A resistor, R, was connected to a circuit box as shown in Fure P 5.4-8. The resistance was changed, and the voltage was measured again. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ.

### Dorf homework solutions

#### Dorf homework solutions

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