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Okay, so I'm working linearly through Fundamentals of Electric Circuits and this is a practice problem.
Obtain the node voltages in the circuit.
http://i2.minus.com/iJPUlZbuy0mLa.jpg
The answer is -6 volts for node 1 and -42 volts for node 2. (Which I checked in a simulator.)
Not really... Well, i = v/R I guess.
Okay, the first step is looking at each node and labelling currents going in and out and using Kirchhoff's current law: the sum of currents flowing into a node equals the sum of currents flowing out of that same node. I am under the impression that I can choose the directions of unknown currents freely, and that if I got them the wrong way then I'll just get a negative current.
So, why not say that for node 1, current flows in from the source and in from the 6 Ω resistor, and out from the 2 Ω resistor? (→↓←) So 3 + (1/6)v2 + (1/6)v1 = (1/2)v1, or 3 = (1/2 - 1/6)v1 + (-1/6)v2. Meanwhile, for node 2, current flows out towards the source, out towards the 6 Ω resistor, and in from the 7 Ω resistor. (←↑→) It works out to 12 = (1/6)v1 + (1/7 - 1/6)v2.
The section introduces using Cramer's method to solve for the unknown voltages here, so I did just that and got 97.2 volts and 176.4 volts. Which obviously is not -6 and -42.
Repeating the process varying the directions of the currents gave different answers. 1:(→↑→) 2:(→↑→) gave 54 and -126. 1:(→↑→) 2:(→↓→) gave 8.18 and -34.36. 1:(→↓→) 2:(→↑→) gave 162 and 630.
Only 1:(→↓→) 2:(→↓→) gives the correct answer -6 and -42. So, either the directions of currents can't be chosen freely, or I'm somehow fumbling what should be simple maths. In the first case, what are the basis of choosing a direction that I'm missing? In the second case, what am I doing wrong?
Thank you for reading!
Homework Statement
Obtain the node voltages in the circuit.
http://i2.minus.com/iJPUlZbuy0mLa.jpg
The answer is -6 volts for node 1 and -42 volts for node 2. (Which I checked in a simulator.)
Homework Equations
Not really... Well, i = v/R I guess.
The Attempt at a Solution
Okay, the first step is looking at each node and labelling currents going in and out and using Kirchhoff's current law: the sum of currents flowing into a node equals the sum of currents flowing out of that same node. I am under the impression that I can choose the directions of unknown currents freely, and that if I got them the wrong way then I'll just get a negative current.
So, why not say that for node 1, current flows in from the source and in from the 6 Ω resistor, and out from the 2 Ω resistor? (→↓←) So 3 + (1/6)v2 + (1/6)v1 = (1/2)v1, or 3 = (1/2 - 1/6)v1 + (-1/6)v2. Meanwhile, for node 2, current flows out towards the source, out towards the 6 Ω resistor, and in from the 7 Ω resistor. (←↑→) It works out to 12 = (1/6)v1 + (1/7 - 1/6)v2.
The section introduces using Cramer's method to solve for the unknown voltages here, so I did just that and got 97.2 volts and 176.4 volts. Which obviously is not -6 and -42.
Repeating the process varying the directions of the currents gave different answers. 1:(→↑→) 2:(→↑→) gave 54 and -126. 1:(→↑→) 2:(→↓→) gave 8.18 and -34.36. 1:(→↓→) 2:(→↑→) gave 162 and 630.
Only 1:(→↓→) 2:(→↓→) gives the correct answer -6 and -42. So, either the directions of currents can't be chosen freely, or I'm somehow fumbling what should be simple maths. In the first case, what are the basis of choosing a direction that I'm missing? In the second case, what am I doing wrong?
Thank you for reading!
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