Kirchhoff's Laws apply the Law of Conservation of Energy and the Law of Conservation of Charge. Kirchhoff's Laws deal with current and voltage in electrical connections. There are two basic types of connections (circuits).
Series Circuit
 electrons flow along one path only:
Parallel Circuit
 electrons flow along more than one pathway(i.e. alternate branches for current to follow. The total current I_{t} will split into branch 1  I_{1} and branch 2  I_{2}):
Kirchhoff's Current Law states….
At any junction in an electric circuit, the total current flowing into the junction is the same as the total current leaving the junction.
Thus …
 For a series circuit the current at all points will be the same since electrons can flow along only one path.
 For a parallel circuit the total current flowing into a connection must equal the sum of the currents flowing out of the connection.
Kirchhoff's Voltage Law states
The algebraic sum of the potential difference around any closed path or loop must be zero.
Thus
 For a series connection the total potential difference is equal to the sum of the potential differences across each component.
V_{t} = V_{1} + V_{2} + V_{3}+....+V_{n}
In a circuit, potential difference increases across the terminals of a cell(power supply) and decrease across the electrical components(i.e. resistors, lamps, speakers,..etc.). The increases in potential difference across the terminals must equal the sum of the decrease in potential difference across each component connected in series. This is a result of the Law of Conservation of Energy.
 For parallel connections the drop in potential difference across all branches are equal.
V_{t} = V_{1} = V_{2} = V_{3}=....V_{n}
Kirchhoff's Laws can also be applied to a circuit which is a combination of a series and a parallel connection.
For example:
Find I_{1}, I_{3}, R_{1}, R_{2}, R_{3}, V_{1}, and V_{2}
Solution
 R_{1} is in series with the loop and therefore the current passing through R_{1} is the same amount entering the loop I_{t}= I_{1} = 620 mA.
 We don't yet know V_{1} so we can not calculate R_{1}.
 620 mA enters the parallel connection, 220 mA travels along one path through R_{2}.
620 mA  220 mA = 400 mA travels along the other path through R_{3}.
Therefore I_{3} = 400 mA
 We know the current passing through R_{3} and the voltage drop accross it, now we can calcuate the resistance.
R_{3} = V_{3} / I_{3} R_{3} = 1.9 V / 0.400 A R_{2} = 4.75 .
 Since the potential difference across the parallel connection is 1.9 V, V_{2} = 1.9 V.
 We know the voltage drop accross R_{2} and the current passing through it, now we can calcuate the resistance.
R_{2} = V_{2} / I_{2} R_{2} = 1.9 V / 0.220 A R_{2} = 8.64 .
 In the loop containing R_{1} and the 2nd loop, the sum of the potential difference drops across the components must equal the potential difference increase. Therefore 4.5 V = 1.9 V + V_{1} and V_{1} = 2.6 V.
 Now we can calculate R_{1}.
R_{1} = V_{1} / I_{1} R_{1} = 2.6 V / 0.620 A R_{1} = 4.19 .
