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Kirchoff's Laws
---------------
Following the conventional current flow that electric charge moves from the
positive side of the battery to the negative side, the rule for voltage
sources is as follows:
. - to + causes an increase in potential.
. + to - causes a decrease in potential.
For resistors (impedances) the rules are as follows:
. Subtract voltage drop in the direction of the current flow.
. Increase voltage in the direction against the current flow.
Consider:
Current Law:
ΣI_{k} = 0 => I_{3} = I_{1} + I_{2}
^{k}
In this case, the direction of the current flows is chosen in accordance
with the conventional current flow, and is consistent for each loop. Thus,
the direction of I_{1} and I_{2} set to be away from the + terminals.
Use B as common node (ground if you like), thus V_{B} = 0
Starting at B and proceeding clockwise around loop BEFAB:
-I_{3}R_{3} - I_{1}R_{1} + V_{1} = 0
V_{1} = I_{3}R_{3} + I_{1}R_{1}
Starting at B and proceeding clockwise around loop BCDEB:
-V_{2} + I_{2}R_{2} + I_{3}R_{3} = 0
V_{2} = I_{2}R_{2} + I_{3}R_{3}
What happens if we start at B and, instead, proceed counter clockwise
around BEDCB:
-I_{3}R_{3} - I_{2}R_{2} + V_{2} = 0
V_{2} = I_{3}R_{3} + I_{2}R_{2} ... the same as the clockwise result
We could also use loops BEFAB or BCDEB and BCDEFAB for the analysis.
Again, we would get the same result. For the loop BCDEFAB we would get,
-V_{2} + I_{2}R_{2} - I_{1}R_{1} + V_{1} = 0
V_{1} - V_{2} = I_{1}R_{1} - I_{2}R_{2}
As long as the rules and current conventions are followed, Kirchoff's
laws work regardless of the chosen loop and direction. The actual net
current directions are reflected in the sign of the answer.
Ex. 1
Find currents through each battery with switch closed and switch open.
Switch closed:
I_{3} = I_{2} - I_{1} (direction of I_{1} and I_{3} set to be away from + terminals)
Starting at B and proceeding clockwise around loop BEFAB:
-9 + 4I_{3} - 2I_{1} + 6 = 0
3 = 4I_{3} - 2I_{1}
3 = 4I_{2} - 6I_{1} .... (1)
Starting at B and proceeding clockwise around loop BCDEB:
-5I_{2} - 4I_{3} + 9 = 0
9 = 5I_{2} - 4I_{3}
9 = 9I_{2} - 4I_{1} .... (2)
From (1) and (2),
12 = 16I_{2} - 24I_{1}
54 = 54I_{2} - 24I_{1}
-42 = - 38I_{2}
I_{2} = 1.105 A
From (1),
3 = 4 * 1.105 - 6I_{1}
I_{1} = (3 - 4.421)/-6 = 0.24 A
and
I_{3} = I_{2} - I_{1} = 1.105 - 0.24 = 0.865 A
Switch open:
I_{2} = 0 so I_{3} = -I_{1}
-9 + 4I_{3} - 2I_{1} + 6 = 0
-3 = 4I_{1} + 2I_{1}
I_{1} = -3/6 = 0.5 A
Ex 2.
I_{3} = I_{1} + I_{2} (direction of I_{1} and I_{2} set to be away from + terminals)
Starting at B and proceeding clockwise around loop BEFAB:
-40I_{3} + 10 - 10I_{1} = 0
10 = 10I_{1} + 40I_{3}
10 = 10I_{1} + 40I_{1} + 40I_{2}
10 = 50I_{1} + 40I_{2}
1 = 5I_{1} + 4I_{2} .... (1)
Starting at B and proceeding clockwise around loop BCDEB:
20I_{2} - 20 + 40I_{3} = 0
20 = 20I_{2} + 40I_{3}
20 = 20I_{2} + 40I_{1} + 40I_{2}
20 = 60I_{2} + 40I_{1}
1 = 2I_{1} + 3I_{2} ....(2)
From (1) and (2),
6 = 30I_{1} + 24I_{2}
8 = 16I_{1} + 24I_{2}
-2 = 14I_{1}
I_{1} = -0.143 A
From (1),
1 = 5I_{1} + 4I_{2} .... (1)
1 + 0.714 = 4I_{1}
I_{2} = 0.429 A
and
I_{3} = I_{1} + I_{2} = -0.143 + 0.429 = 0.286 A