- #1
wellcoughed
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Hi,
Ive trying to work through Kirchhoffs voltage and current laws.
Ive so far calculated:
Current in Loop one (I1): 3 Amps going anticlockwise
Current in Loop Two (I2): 4 amps going anticlockwise
Current in Loop Three (I3): 8 amps going anticlockwise
R1: 5 amps going up = (Current in loop three - current in loop one)
-5amps going down = (current in loop one - current in loop three)
R2: 3 amps going up
-3 amps going down
R3: 4 amps going up = (Current in loop three - current in loop two)
-4 amps going down = (current in loop two - current in loop three)
R4: 4 amps going up
-4 amps going down
R5 -1 amps left to right = (current in loop one - current in loop two)
1 amps right to left = (current in loop two - current in loop one)
All these currents are going anticlockwise but the problem they've given me has 8 amps written on it going clockwise.
So when I apply Kirchhoffs Current Laws (The sum of currents going into a node plus the currents going out of a node = 0) starting with the 8amps clockwise they've given me I find the opposite to what I've found when I solved the voltage laws.
Specifically using the current laws starting with 8amps clockwise I now have:
R1: 5 amps going down the resistor
R2: 3 Amps going down the Resistor
R3: 4 Amps going down the resistor
R4: 4 Amps going down the Resistor
R5: 1 Amps going left to right
Am I right in saying that By using Kirchoffs voltage laws I've found the electron flow. The direction the electrons flow from the negative of the battery to the positive.
But if I use kirchhoffs current laws and start with the "8 Amps clockwise" the lecturer has given me in the problem I am now actually finding the direction of the conventional flow?
Which is why it's directly opposite to the electron flow I've found when I used Kirchhoffs voltage laws?
many thanks if you could clear this up :)
Ive trying to work through Kirchhoffs voltage and current laws.
Ive so far calculated:
Current in Loop one (I1): 3 Amps going anticlockwise
Current in Loop Two (I2): 4 amps going anticlockwise
Current in Loop Three (I3): 8 amps going anticlockwise
R1: 5 amps going up = (Current in loop three - current in loop one)
-5amps going down = (current in loop one - current in loop three)
R2: 3 amps going up
-3 amps going down
R3: 4 amps going up = (Current in loop three - current in loop two)
-4 amps going down = (current in loop two - current in loop three)
R4: 4 amps going up
-4 amps going down
R5 -1 amps left to right = (current in loop one - current in loop two)
1 amps right to left = (current in loop two - current in loop one)
All these currents are going anticlockwise but the problem they've given me has 8 amps written on it going clockwise.
So when I apply Kirchhoffs Current Laws (The sum of currents going into a node plus the currents going out of a node = 0) starting with the 8amps clockwise they've given me I find the opposite to what I've found when I solved the voltage laws.
Specifically using the current laws starting with 8amps clockwise I now have:
R1: 5 amps going down the resistor
R2: 3 Amps going down the Resistor
R3: 4 Amps going down the resistor
R4: 4 Amps going down the Resistor
R5: 1 Amps going left to right
Am I right in saying that By using Kirchoffs voltage laws I've found the electron flow. The direction the electrons flow from the negative of the battery to the positive.
But if I use kirchhoffs current laws and start with the "8 Amps clockwise" the lecturer has given me in the problem I am now actually finding the direction of the conventional flow?
Which is why it's directly opposite to the electron flow I've found when I used Kirchhoffs voltage laws?
many thanks if you could clear this up :)
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