Hello, thank you for taking the time to look at my question and help. I am trying to self study in preparation for a course out of a book with mathematical electricity problems. I am stuck on a question involving Kirchhoff's Voltage Law. In this question you are given a circuit and are told to solve for the currents through the resistors using the KVL and simultaneous equations. The way it is explained in the book is like this "emf is considered positive if the negative terminal is met first" and "IR drop is considered positive if the current in resistor R is in the same direction as path being followed"
The yellow arrows are the directions of current assumed and the red ones are the directions of the "path". The resistances by the emfs are internal to the emf and have to be taken in to account, and are considered as in series with the resistors in the branch.
Following these instructions I get two equations like this:
-4v = +(11R)I1 + (8R)I2
-8v = +(1R)I3 + (8R)I2
Since I3 = I1 + I2
-8v = (+1R)I1 + (9R)I2
Multiplying second loop equation by 11 to cancel I1 and adding the two equations.
-4v = -(11R)I1 + (8R)I2
-88v = +(11R)I1 + (99R)I2
-92 = (107)I2
-92/107= I2 = -0.859 < Not correct. Neither magnitude or polarity.
The correct I2 = 0.785 to the left, I1 = 0.935 to the left.
I noticed that if the polarity of the 88 volts were positive then in the end we would get
84/107 = I2 = 0.785 and we can derive the rest of the currents substituting I2 in the equation.
I feel as though I followed all the instructions so I don't know why I cant get 0.785 as I2.
Can anyone help? Much thanks!
The yellow arrows are the directions of current assumed and the red ones are the directions of the "path". The resistances by the emfs are internal to the emf and have to be taken in to account, and are considered as in series with the resistors in the branch.
Following these instructions I get two equations like this:
-4v = +(11R)I1 + (8R)I2
-8v = +(1R)I3 + (8R)I2
Since I3 = I1 + I2
-8v = (+1R)I1 + (9R)I2
Multiplying second loop equation by 11 to cancel I1 and adding the two equations.
-4v = -(11R)I1 + (8R)I2
-88v = +(11R)I1 + (99R)I2
-92 = (107)I2
-92/107= I2 = -0.859 < Not correct. Neither magnitude or polarity.
The correct I2 = 0.785 to the left, I1 = 0.935 to the left.
I noticed that if the polarity of the 88 volts were positive then in the end we would get
84/107 = I2 = 0.785 and we can derive the rest of the currents substituting I2 in the equation.
I feel as though I followed all the instructions so I don't know why I cant get 0.785 as I2.
Can anyone help? Much thanks!