HW _8 Solutions - Problem 3.67 Determine the maximum power...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 3.67 Determine the maximum power that can be extracted by the load resistor from the circuit in Fig. P3.67. 2000 I x 6 k Ω 3 k Ω 4 k Ω R L I x 15 V + _ _ Figure P3.67: Circuit for Problem 3.67. Solution: To find the Th´evenin equivalent circuit, we start by determining V Th = V oc . 2000 I x 6 k Ω 3 k Ω 4 k Ω I x 15 V _ _ V oc a b V 1 + _ Voltage division: V 1 = 15 ( 3 + 6 ) k × 6k = 10 V I x = V 1 6k = 10 6 mA . The dependent voltage source is: 2000 I x = 2 × 10 6 × 10 3 × 10 3 = 20 6 V . With ( a , b ) an open circuit, no current flows through the 4-k resistor. Hence, there is no voltage drop across it. V Th = V oc = V 1 2000 I x = 10 20 6 = 40 6 = 6 . 67 V . 2000 I x 6 k Ω 3 k Ω 4 k Ω 15 V _ _ I sc a b I 1 I 2 Next, we find I sc : 15 + 3k I 1 + 6k ( I 1 I 2 ) = 0
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
6k ( I 2 I 1 )+ 4k I 2 + 2000 I x = 0 Also, I x = I 1 I 2 Solution yields: I 1 = 2 . 5 mA , I 2 = 1 . 25 mA . I
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/15/2011 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

Page1 / 7

HW _8 Solutions - Problem 3.67 Determine the maximum power...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online