Circuits can be in series or parallel.

As the name suggests, loads that appear in sequence on the one circuit branch are considered to be in series.

Voltages and resistances in series can be added together and treated as a single voltage or resistance:

$$V_1 + V_2 + ..... V_n = V_{total}$$where n is the total number of voltages in series.

$$R_1 + R_2 + ..... R_n = R_{total}$$where n is the total number of resistances in series.

Buildings do not usually arrange electrical devices and lights in series. If they did, if one device were switched off, all the current in the single circuit would be stopped, and all the devices would switch off!

To solve this problem, lights and other electrical equipment in buildings are arranged in parallel.

A circuit may divide into two or more branches, each of which have the same voltage.

Resistances in parallel can be added together and treated as a single resistance according to the following formula:

$$1/{R_{total}} = 1/{R_1} + 1/{R_2} + ..... 1/{R_n} $$where n is the total number of resistances in series.

Batteries may also be arranged in parallel. However, this has no effect on the voltage of the circuit, or the current. It does, however, provide the circuit with more energy. Lights will be the same brightness, but will stay on longer with two batteries instead of one.

In the diagram, there are three resistors of different resistances in parallel. The voltage across all three is the same, and is equal to the potential across the electrodes of the battery: 6.0V.

The currents through the individual branches of the circuit may be calculated from Ohm's Law:

$$I = V/R$$Resistor | Voltage | Current |
---|---|---|

R1 2.0Ω | 6.0 V | 3.0 A |

R2 3.0Ω | 6.0V | 2.0 A |

R3 4.0Ω | 6.0V | 1.5 A |

The total current is the sum of the individual branches: $I_{total} = 3 + 2 + 1.5 = 6.5A$.

To check this, let us use the parallel resistance formula to calculate the effective resistance across the block of three resistors:

$$1/{R_{total}} = 1/{R_1} + 1/{R_2} + 1/{R_3} = 1/2 + 1/3 + 1/4 = {(6+4+3)}/{12} = {13}/{12}$$The effective resistance across the 3 resistances is therefore ${12}/{13} = 0.923Ω$

Using this to calculate the current through the whole circuit:

$I = V/R = 6/0.923 = 6.5A$

This result is the same as adding up the three currents individually.

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