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Chapter 18
Kirchhoff’s Rules
RC Circuits
Junction Rule
1) The sum of the currents entering a junction
must equal the sum of currents leaving a
junction. (Conservation of Charge)
Loop Rule
2) The sum of potential differences across all
the elements on any closed loop in a circuit
must be zero. (Conservation of Energy)
I1
I
3W
5W
I
10W
2W
I2
I3
2W
V
The Junction Rule says that
I = I1 + I 2 + I3
I1
I
3W
5W
I
10W
2W
I2
I3
2W
V
We can use the loop rule multiple times:
V - 3 I - 10 I2 - 2I = 0
V - 3I - 2I3 - 2I = 0
V - 3 I - 5 I1 - 2I = 0
We now have a system of 4 equations and 4
unknowns. We can therefore solve for each
of the 4 unknown currents.
V - 3 I - 10 I2 - 2I = 0
V - 3I - 2I3 - 2I = 0
I = I1 + I2 + I3
V - 3 I - 5 I1 - 2I = 0
V  5I
I2 
10
Plug these three into
the current equation.
V  5I
I3 
2
V  5I
I1 
5
V  5I V  5I V  5I
I 


5
10
2
10I = 8V - 40 I
I = 0.16 V
Plug this back into the other 3 equations...
I1 = 0.04 V
I2 = 0.02 V
I3 = 0.1 V
For Applying Kirchhoff’s Rules
A) assign a direction to the current in each
branch of the circuit. Just GUESS!! If your
guess is incorrect, the current will come out
as a negative number, but the magnitude will
still be correct!
B) when applying the loop rule, you must
choose a consistent direction in which to
proceed around the loop (either clockwise or
counterclockwise, your choice, but stick to it).
1) When you encounter a resistor in the direction
of the current, the voltage drop is DV = - I R
2) When you encounter a resistor opposite the
current, the voltage drop is DV = + I R
I
- I1R1
I2
R1 I1
R2
+I2R2
3) When you encounter an emf in the direction
you’re going around the loop, the voltage change
is +DV
4) When you encounter an emf opposite the
the direction you’re going, the voltage change
is -DV
- V1
V1
V2
+ V2
5) The junction rule can only be applied n-1
times in a circuit with n junctions.
6) Each new equation you write must contain
a current that you haven’t yet used.
7) To solve a system of equations with k
unknown quantities, you need k independent
equations.
I1
I
3W
5W
I
10W
2W
I2
I3
2W
V
This circuit has two junction points. Therefore,
we can only use the junction rule ONE time.
I0
+
18 V
d
e
I2
6W
I1
+
a
1W
b
_
12 W
12 V
_
c
1W
f
Find the currents in each branch of this circuit.
There are two junction points, so we can apply
the junction rule ONCE.
I0 = I1 + I2
12 W
18 V
e
6W
+
+
1W
_
a
b
12 V
_
d
c
1W
f
How many possible loops are there?
bcda
efcdab
efcb
We have 3 unknowns, so we need 3 equations
total. Therefore, we need use only 2 of the 3
equations provided by the loop rule. (The junction
rule gave us 1 equation already!)
12 W
+
18 V
d
I2
6W
I1
+
I0
e
_
a
1W
b
12 V
_
c
1W
f
Let’s go around bcda first.
-(6W) I1 + 18 V - (12W) I0 = 0
(12W) I0 + (6W) I1 = 18 V
12 W
+
18 V
d
I2
6W
I1
+
I0
e
_
a
1W
b
12 V
_
c
1W
f
Let’s go around efcb second.
12 V - (1W) I2 + (6W) I1 - (1W) I2 = 0
(2W) I2 - (6W) I1 = 12 V