1. Lecture 24
Chapter 18
Kirchhoff’s Rules
RC Circuits
Monday
22 March 1999
Physics 112

2. The Physics 112
Help Session
Tuesdays
6:00 - 7:00 pm
CANCELLED!
Mondays
5:30 - 7:00 pm
NSC Room 118

3. Kirchhoff's Rules Junction Rule Loop 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)

4. 2W
10W 5W 3W I I3 I2 I1 V
The Junction Rule says that
I = I1 + I2 + I3

5. 2W
10W 5W 3W I I3 I2 I1 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

6. 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
V - 3 I - 5 I1 - 2I = 0
I = I1 + I2 + I3
Plug these three into
the current equation.

7. 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

8. For Applying Kirchhoff’s Rules
Additional Notes
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).

9. - I1R1 +I2R2 Still More Notes!
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 R2 I R1 I1 I2
- I1R1
+I2R2

10. You Guessed It!
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

11. Last Page of Notes 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.

12. 2W
10W 5W 3W I I3 I2 I1 V
This circuit has two junction points. Therefore,
we can only use the junction rule ONE time.

13. I0 = I1 + I2 Find the currents in each branch of this circuit.
6 W
1 W
18 V
12 W
12 V + _ a f e c b d I2 I0 I1
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

14. 3 How many possible loops are there? b c d a e f c d a b e f c b
18 V
12 W
12 V + _ a f e c b d 3
How many possible loops are there?
b c d a
e f c d a b
e f c b
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!)

15. Let’s go around bcda first.
6 W
1 W
18 V
12 W
12 V + _ a f e c b d I2 I0 I1
Let’s go around bcda first.
-(6W) I V - (12W) I0 = 0
(12W) I0 + (6W) I1 = 18 V

16. Let’s go around efcb second.I0
6 W
1 W
18 V
12 W
12 V + _ a f e c b d I1 I2
Let’s go around efcb second.
12 V - (1W) I2 + (6W) I1 - (1W) I2 = 0
(2W) I2 - (6W) I1 = 12 V