HW _8 Solutions

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

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

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

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

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online