1. Analysis with Dependent Sources
Topic 6
Analysis with Dependent Sources
(2.5)

2. Look for interim objectives+ -
500v 5Ω
20Ω
5iΔ iΔ vo io a b c
Find vo in the circuit
Look for interim objectives
If we know io then we can find vo by Ohm’s Law
By KVL: 2
If we know iΔ then we can find 5 iΔ (and also io).
Eliminating io: 3 So
& 4
By KCL:
1 into 4 5 So 1
By Ohm’s Law: 6
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Analysis with Dependent Sources
Where’s the other 20 volts?

3. Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed.
Assessment 2.9
30i1 a b c d e f
Let’s label all the nodes + - 5v
54kΩ
6kΩ i1 v
30i1 1v
1.8kΩ 8v i1
30i1 i2 i1
i1 runs through the 5v and 1v sources as well as the 54kΩ resistor
Similarly, 30i1 runs through the 1.8kΩ resistor and the 8v source.
By KCL at node c: So
By KVL around loop a-b-c-f-a:
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Analysis with Dependent Sources

4. Analysis with Dependent Sources
Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed.
Assessment 2.9 + - 5v
54kΩ
6kΩ i1 v
30i1 1v
1.8kΩ 8v a b c d e f i2
Writing KVL around loop e-d-c-f-e
4/30/2019
Analysis with Dependent Sources

5. Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed.
Assessment 2.9 + - 5v
54kΩ
6kΩ i1 v
30i1 1v
1.8kΩ 8v a b c d e f i2
(c)
The dependent source is not generating
Source
Other
Power 5v
25µA
125µW 1v
25µW v
750µA
-1500µW 8v
6000µW
total generated = 6150µW
4/30/2019
Analysis with Dependent Sources

6. Analysis with Dependent Sources
Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed.
Assessment 2.9 + - 5v
54kΩ
6kΩ i1 v
30i1 1v
1.8kΩ 8v a b c d e f i2
(d)
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Analysis with Dependent Sources

7. Assessment 2.10 2iφ i1 a The current iφ is 2A. Calculate vs. b 10Ω iφ
30Ω
10Ω vs iφ
2iφ i1
The current iφ is 2A. Calculate vs. a b
Label the nodes
Define a current, i1, through the 10Ω resistor c
KCL at node a: 1 2
Solving for i1:
KVL around loop b-a-c-b: 3 4
Solving for vs:
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Analysis with Dependent Sources

8. Assessment 2.10 iφ=2A 2iφ i1=1A vs=70v
30Ω
10Ω vs iφ
2iφ
(b) Calculate the power absorbed by the independent voltage source. i1 a b is
To calculate power absorbed let’s define a downhill current, is c
Then power absorbed by the independent voltage source is
So power absorbed by the independent voltage source:
Writing KCL at b:
Solving for is:
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Analysis with Dependent Sources

9. Assessment 2.10 iφ=2A 2iφ i1=1A vs=70v is=3A
30Ω
10Ω vs iφ
2iφ
(c) Calculate the power delivered by the independent current source. i1 a b is vi + -
To calculate power delivered let’s define a voltage, vi, so the current is uphill c
For an uphill current, power delivered is:
To find vi, write KVL around the lefthand loop c-a-c: So
&
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Analysis with Dependent Sources

10. Assessment 2.10 vφ + - 5A
30Ω
10Ω vs iφ
2iφ
(d) Calculate the power delivered by the controlled current source. i1 a b is vi + -
Again define voltage, vφ so the current is uphill c
For an uphill current, power delivered is:
iφ=2A
i1=1A
vs=70v
is=3A
vi=60v
Note that vφ also appears across the 10Ω resistor (KVL)
So, by Ohm’s Law:
Giving power delivered by the controlled current source:
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Analysis with Dependent Sources

11. Analysis with Dependent Sources
Assessment 2.10 vφ + - 5A
30Ω
10Ω vs iφ
2iφ
(e) Calculate the total power absorbed by the two resistors. i1 a b is vi + -
Since we know the currents we’ll just use i2R c
iφ=2A
i1=1A
vs=70v
is=3A
vi=60v
Power check
4/30/2019
Analysis with Dependent Sources