The document explains Kirchhoff's rules, which are used to solve complex electrical circuits through the principles of conservation of charge and energy. It details terminology such as loops and junctions, and describes the application of the junction rule, where the sum of currents entering a junction equals the sum leaving, and the loop rule, which states that the sum of potential changes around any closed circuit path must be zero. Additionally, it provides guidance on tracing current direction and applying Ohm's law.

1. Kirchoff’s Rules
Purpose: solve complex circuits

2. Some terminology
• Loop: ANY closed
5Ω 12 V
path in a circuit. This
one has three loops.
The loop direction is 10 Ω
arbitrary.
Best practice: follow out
of the + side of
battery when 20 Ω
6V
possible.
Junction: where two or
more wires meet.

3. Purpose of Rules:
• Since some circuits are too complex they
must be solved using Kirchoff’s Rules.
These rules follow the principle and ideals
of conservation of charge and
conservation of energy.
The rules help keep track of +/- for
potentials.

4. Before applying rules, label
currents in loops
Choose current directions. 5Ω 12 V
Use arrows right on the
wire to indicate current
and label with I1 I2, and/or
10 Ω
I3.
Cannot have all three
currents enter or leave a
junction simultaneously.
At least one direction 20 Ω
6V
must be different coming
in or out of a junction.

5. 1. Junction Rule
• At any junction the sum
of all currents entering
must equal the sum of all
currents leaving the 5Ω 12 V
junction.
This is conservation
10 Ω
Of charge.
Equations:
20 Ω
6V

6. Loop Rule
• The sum of the
changes in potential 12 V
5Ω
around any closed
path of a circuit must
be zero. 10 Ω
This is conservation of
energy.
20 Ω
6V

7. Voltage: + or - ???
• When tracing a current
5Ω 12 V
through a loop, start at
a battery.
The V will be + when you 10 Ω
trace neg to pos. (uphill)
The V will be – when you
trace pos to neg. 20 Ω
6V
(downhill)

8. IR values through resistors
• Apply Ohms Law:
5Ω 12 V
V = IR
IR will be neg when
direction of loop and 10 Ω
current match.
IR will be pos when loop
trace goes against 20 Ω
6V
current

9. The idea is to get all the equations
and the correct signs
5Ω 12 V
• Apply junction rule for
current equation
10 Ω
• Apply loop rule and
sign protocol for IR
and battery voltages
20 Ω
6V

10. The idea is to get all the equations
and the correct signs
5Ω 12 V
• Apply junction rule for
current equation
10 Ω
• Apply loop rule and
sign protocol for IR
and battery voltages
20 Ω
6V