sevag00
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Homework Statement
Find Vx, Is and power on the dependent source.
The Attempt at a Solution
http://s17.postimg.org/3x4hszzkf/002.jpg
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Finding Vx, Is, and Power in a Dependent Source
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Find Vx, Is and power on the dependent source.
http://s17.postimg.org/3x4hszzkf/002.jpg
Your determination of vx is fine. However, Is is not the only current flowing through the 4 Ω resistor; there's some current coming from the 29 A source, flowing through the 4 Ω + Is combination, then to the components on the right.
You should be in a position to know how much current is entering and leaving that middle bit.
You should be in a position to know how much current is entering and leaving that middle bit.
sevag00
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I did KCL on node a and got 30 A. Also calculated the current through the 4 Ω resistor -1.5 A.
Then KCL on node b, Is = 31.5 A
Then KCL on node b, Is = 31.5 A
sevag00
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Oh ok. 28 not 30. Then Is = 26.5.
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sevag00
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sevag00 said:
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sevag00
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Here's the new sketch. If I'm not wrong then 29.5 A.
sevag00
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sevag00
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Another day, Another KVL-KCL problem.
Find V0 using KVL-KCL and Ohm's law.
Here's my solution
http://s21.postimg.org/usa92r31z/001.jpg
Find V0 using KVL-KCL and Ohm's law.
Here's my solution
http://s21.postimg.org/usa92r31z/001.jpg
Last edited by a moderator:
You really should place new questions in new threads, if for no other reason than more people are likely to see it.
For this problem it looks like you've got the right ideas, but you're losing way too much accuracy by not keeping enough decimal places in your intermediate results. Keep intermediate values in fraction form or keep several more decimal places ("guard digits").
For example, knowing that I1 is 4 mA and it's flowing through a 4 kΩ resistor you can tell that the potential drop across that resistor should be 16 V exactly. That means the sum of the 42V supply and the drop across the 6 kΩ resistor must add to 16V, so that the drop is 26 V across that resistor. But in your determination of i2 you rounded the result to 4.3 mA, and then when you calculated the potential drop across that resistor you arrived at 25.8 V.
The situation gets worse when you round small values, as the resulting percentage change is more drastic. You determined the current through the 18 kΩ resistor to be 1.63 mA, but it should have ended up as 1.889 mA. So the early roundings of values has landed you with a 14% error so far...
For this problem it looks like you've got the right ideas, but you're losing way too much accuracy by not keeping enough decimal places in your intermediate results. Keep intermediate values in fraction form or keep several more decimal places ("guard digits").
For example, knowing that I1 is 4 mA and it's flowing through a 4 kΩ resistor you can tell that the potential drop across that resistor should be 16 V exactly. That means the sum of the 42V supply and the drop across the 6 kΩ resistor must add to 16V, so that the drop is 26 V across that resistor. But in your determination of i2 you rounded the result to 4.3 mA, and then when you calculated the potential drop across that resistor you arrived at 25.8 V.
The situation gets worse when you round small values, as the resulting percentage change is more drastic. You determined the current through the 18 kΩ resistor to be 1.63 mA, but it should have ended up as 1.889 mA. So the early roundings of values has landed you with a 14% error so far...
sevag00
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Is it enough if i write three decimal digits?