**When to
use Kirchhoff's laws**

Here are three typical circuit diagrams that might need to be solved (e.g. given the resistances of all the resistors and the voltages of all the batteries, find all of the currents).

The figure on the left
can be solved easily using the
rules for adding resistances in parallel and series.
Resistors **R**_{3},
**R**_{4}
and **R**_{5}
can be treated as if they were a single resistor of resistance
**R**_{345}.
This effective resistance can be added in parallel with
**R**_{2}
to give another effective resistance
**R**_{2345}.
Finally, that effective resistance can be added in series with
**R**_{1}
to calculate the net resistance
**R = R**_{1}**
+ R**_{2345} seen by the
battery. This value of R can be used to compute the current produced by the
battery. The current from the battery all flows through
**R**_{1}, so
the voltage drop across
**R**_{1}
can be computed using Ohm's law. Since the battery voltage is known, you can
next compute the voltage drop across
**R**_{2}. Now
applying Ohm's law to
**R**_{2}
tells you the current through it; etc.
In this way you eventually figure out the currents and voltage drops
everywhere in the circuit.

The middle and right-side figures cannot be solved by such a simple(?) strategy. Instead, you must write down Kirchhoff's laws and solve the equations.

For
instance consider the right-hand figure. Give names to the
unknown currents
i_{1},
i_{2}
and i_{3}
through the 3 resistors (choosing the directions ad lib).
These unknown currents can be found by solving 3
equations: |

- The first equation describes current conservation into the node.
- The second equation expresses the requirement that voltage losses cancel voltage gains for the loop on the left.
- The third
equation expresses the same for the loop on the right.
Note that if one makes a counter-clockwise loop,
one goes against the current through
**R**_{3 }and the sign of that term is therefore opposite the normal one.

You could also write an equation for a third loop that is defined by traveling around the outside of the circuit. That equation would be slightly more complicated because it would involve both batteries; but it is unnecessary since it would be merely a linear combination of Eq. 2 and Eq. 3 above.