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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 nodevoltage 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 nodevoltage 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 nodevoltage equation:
va − vb
1
va − vb
vb
+
c
− vb =
R3
R5
R3
R4 Along with the ﬁrst nodevoltage 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...
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This document was uploaded on 04/05/2014.
 Fall '09
 Volt

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