- #1
wcjy
- 73
- 10
- Homework Statement:
- Determine the current I (in Amperes) in the circuit below using nodal analysis.
- Relevant Equations:
- -
Ans: 0.123A
Tried mesh analysis and got it but didnt for nodal analysis
Last edited:
40I is from the current controlled voltage source. Isn't 40I in volts?40I-V0? Subtracting a voltage from a current?
The rotated square (diamond) symbol with '+ -' is a controlled voltage source.40I-V0? Subtracting a voltage from a current?
Can't see a problem. I also get I = 0.112(335)A and (for information) = 5.28(634) V. If you plug these values into check, you find that they give correctly balanced currents. Are you certain the answer should be 0.123A?
the system said the correct answer was 0.123. But anyways thanks, many people got 0.112 also. probably the answer is wrong. Gonna check with my professor.Can't see a problem. I also get I = 0.112(335)A and (for information) = 5.28(634) V. If you plug these values into check, you find that they give correctly balanced currents. Are you certain the answer should be 0.123A?
You could post your mesh analysis for checking. But purple on a black background is pretty eye-unfriendly and probably puts a lot of people off. (Typed-up using Latex is the preferred method.)
Ok, I'd never come across that concept.The rotated square (diamond) symbol with '+ -' is a controlled voltage source.
https://1.bp.blogspot.com/-VWAqITUrvhA/UY-k_NBxwfI/AAAAAAAAAUM/BPVlWJIFgXQ/s1600/sources.png
You can tell it's a voltage source because the '+ -' indicates polarity (as opposed to an arrow for a current source indicating current's direction).
'40I' indicates the output-voltage is numerically (in volts) 40 times the value of the current labelled 'I' (in amps) on the diagram.
40I-V0? Subtracting a voltage from a current?
Yes. Also, it's in series with a resistor, which doesn't make a lot of sense for a current source.You can tell it's a voltage source because the '+ -' indicates polarity (as opposed to an arrow for a current source indicating current's direction).
True, but by convention the units are usually omitted from the diagram and are assumed to be appropriate for the "conversion" implied by the sensed value and resulting output of the controlled source.Ok, I'd never come across that concept.
But the correct way to specify its value would be with units, namely, as 40IΩ.
yea the answer is wrong!This is not something covered on the courses I teach but I was intrigued so decided to try to find out how to do such problems and have a go.
Perhaps not surprisingly, I can't get 0.123A as an answer. However, I'm wary of suggesting that's because the answer is wrong until I know for sure.
Please, @wcjy - it would be great if you could confirm one way or another when you ask your prof. Thanks :)