Finally in circuits with the dependent sources

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: e applied to circuits with dependent sources. Finally, in circuits with the dependent sources, energies may not balance in the manner that we have expected for circuits that we have seen till now. EE 442 dependent sources – 5 Example 1 Find the voltage across R3 in the circuit below. R1 2 Ω VS + 20 V – R3 5 Ω R2 20 Ω R5 2 Ω iR3 R4 10 Ω + – Vd = 8iR3 Use the node-voltage method. Define ground at the bottom. R1 VS EE 442 + – va R2 R3 vb iR3 R4 R5 + – Vd = 8iR3 dependent sources – 6 R1 VS + – va R3 vb iR3 R4 R2 R5 + – Vd = 8iR3 Write the node-voltage equations. (If you need to, fill in the missing steps.) VS − va va va − vb = + R1 R2 R3 va − vb Vd − vb vb + = R3 R5 R4 At this point, we don’t know Vd, so these are two equations in three unknowns. However, we obtain a third equation easily by using the definition for Vd. Vd = c · iR3 EE 442 ￿ va − vb =c R3 ￿ where c = 8 Ω. dependent sources – 7 Inserting the expression for Vd into the second node-voltage equation: ￿￿ ￿ ￿ va − vb 1 va − vb vb + c − vb = R3 R5 R3 R4 Along with the first node-voltage equation, this gives us two equations in the two unknowns. Working out the algebra and inserting the numbers gives: va = 16 V and vb = 10 V. So the voltage across R3 is va – vb = 6 V. EE 442 dependent sources – 8...
View Full Document

This document was uploaded on 04/05/2014.

Ask a homework question - tutors are online