Kirchhoff’s Law

Kirchoff Laws

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).

1. Series Circuit

• electrons flow along one path only:

2. Parallel Circuit • electrons flow along more than one

pathway(i.e. alternate branches for current to follow. The total current It will split into branch 1 - I1 and branch 2 - I2):

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.

• for a series circuit the current at all points will be the same since electrons can flow along only one path.

• It= I1=I2=I3=…….

• For a parallel circuit the total current flowing into a connection must equal the sum of the currents flowing out of the connection.

• It=I1+I2+I3+……….

Kirchhoff’s Voltage Law states

• The algebraic sum of the potential difference (V) around any closed path or loop must be zero.

Series

• For a series connection the total potential difference is equal to the sum of the potential differences across each component.

• Vt = V1 + V2 + V3+....+Vn

Parallel

• For parallel connections the drop in potential difference across all branches are equal.

• Vt = V1 = V2 = V3=....Vn

Series Parallel

Current same add

PotentialDifference(voltage)

add same

• Ex.

Ex.

Recall: Resistance in circuits

• Resistance in a series circuit:– Rt=R1+R2+R3+........

• Resistance in parallel circuit:– 1/Rt= 1/R1+1/R2+1/R3

Series Parallel

Current same add

PotentialDifference(voltage)

add same

Resistance add Add the reciprocal

Series + Parallel

• Kirchhoff's Laws can also be applied to a circuit which is a combination of a series and a parallel connection. For example:

• Find I1, I3, R1, R2, R3, V1, and V2

Combined Circuits

• Some circuits consist of both series and parallel

• To solve these you must find totals for each parallel circuit and then treat the entire circuit as a series circuit

Example

Example

• Find I1, I3, R1, R2, R3, V1, and V2

Solution• R1 is in series with the loop and therefore the current passing

through R1 is the same amount entering the loop It= I1 = 620 mA. • We don't yet know V1 so we can not calculate R1. • 620 mA enters the parallel connection, 220 mA travels along one

path through R2. 620 mA - 220 mA = 400 mA travels along the other path through R3.Therefore I3 = 400 mA

• We know the current passing through R3 and the voltage drop across it, now we can calcuate the resistance.R3 = V3 / I3 R3 = 1.9 V / 0.400 A R2 = 4.75 .

• Since the potential difference across the parallel connection is 1.9 V, V2 = 1.9 V.

• We know the voltage drop accross R2 and the current passing through it, now we can calcuate the resistance.R2 = V2 / I2 R2 = 1.9 V / 0.220 A R2 = 8.64 .

• Practice problems