Hello again,
since help came so quickly with my previous question, I thought I'd give it a new shot
Here http://www.allaboutcircuits.com/worksheets/dcmesh.html I don't understand question 4.
The way I see it: current in the first loop goes clockwise, so polarities of resistors are:
The left-most resistor: left -, right + (since current upstream/enters resistor, so negative on the left side, downstream/exit resistor, positive on the right side). Correct?
Middle resistor (as seen from the first loop mesh current): top -, bottom + (same reason). Is this correct?
In that case, when you'd imaginary place a voltmeter over all the loads and go around the first loop and start at the 6V battery with red lead at the positive terminal and black lead at the negative, you'd get a reading of 6, then I'd move the imaginary meter up over the left most resistor, I'd get a reading of 1000I1 Volts, then to the middle resistor, a reading of 1000(I1-I2) Volts, then on the 1 volt source of course a reading of 1V. These figures would ALL be positive, since the red lead would consequently be attached to a positive terminal (both resistors AND batteries have their polarities in the same, assumed Mesh-current direction, right?)
So how come the KVL equitation given in the answers is: 6 − 1000I1 − 1000(I1 + I2) + 1 = 0 ??
I'm probably seeing this wrong but I compared it to other exercises and to the theory described in the chapter and I can't seem to find what's wrong in my way of thinking here...
Thanks!
S.
since help came so quickly with my previous question, I thought I'd give it a new shot
Here http://www.allaboutcircuits.com/worksheets/dcmesh.html I don't understand question 4.
The way I see it: current in the first loop goes clockwise, so polarities of resistors are:
The left-most resistor: left -, right + (since current upstream/enters resistor, so negative on the left side, downstream/exit resistor, positive on the right side). Correct?
Middle resistor (as seen from the first loop mesh current): top -, bottom + (same reason). Is this correct?
In that case, when you'd imaginary place a voltmeter over all the loads and go around the first loop and start at the 6V battery with red lead at the positive terminal and black lead at the negative, you'd get a reading of 6, then I'd move the imaginary meter up over the left most resistor, I'd get a reading of 1000I1 Volts, then to the middle resistor, a reading of 1000(I1-I2) Volts, then on the 1 volt source of course a reading of 1V. These figures would ALL be positive, since the red lead would consequently be attached to a positive terminal (both resistors AND batteries have their polarities in the same, assumed Mesh-current direction, right?)
So how come the KVL equitation given in the answers is: 6 − 1000I1 − 1000(I1 + I2) + 1 = 0 ??
I'm probably seeing this wrong but I compared it to other exercises and to the theory described in the chapter and I can't seem to find what's wrong in my way of thinking here...
Thanks!
S.