Series and Parallel Circuits – Learn


A series circuit has only one path for electrical current to flow. The electrical components are connected one after another. If one light breaks or a switch is opened in a series circuit the rest of the lights will turn off as there is not a complete circuit. As more lights are added to a series circuit, the lights will get dimmer.

A parallel circuit contains more than one path for electrical current to flow. If one light bulb goes out or a switch is opened, the current can continue to flow along another path. Parallel circuits allow individual components to be switched on and off without impacting other components. Adding or removing light globes does not effect the brightness of each globe.

When a circuit contains more than one resistor, Ohm’s law alone is not sufficient to calculate the current flowing through and the potential difference across each resistor. Circuits with more than one resistor need to be analysed in sections. Calculations will also depend on whether the circuit is in series or parallel.


Analysing series circuits

When resistors are connected in series, the following rules apply:

  • The current through each resistor is the same.
  • The sum of the potential differences across each resistor is equal to the potential difference (voltage) provided to the total circuit. This is also known as Kirchhoff’s voltage law.
  • The total resistance of the circuit is equal to the sum of the individual resistances. This is demonstrated by the following equation and is known as the equivalent effective resistance:

{ R }_{ series }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }....+{ R }_{ n }

Analysing parallel circuits

When resistors are connected in parallel, the following rules apply:

  • The voltage across each resistor is the same.
  • The current is shared between the resistors. This is also known as Kirchhoff’s current law.
  • The inverse of the total resistance of the circuit is equal to the sum of the inverse of the individual resistances. The equivalent effective resistance of a parallel circuit is given by:

\cfrac { 1 }{ { R }_{ parallel } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } ....+\cfrac { 1 }{ { R }_{ n } }

Complex problems may involve circuits that have both series and parallel sections. In these cases, the parallel sections are best dealt with first, followed by the series sections.


Example 1: Series circuit:

a) Calculate the effective equivalent resistance of the circuit:

b) Calculate the current flowing through the circuit:

c) Calculate the potential difference across { R }_{ 2 }:

Answers:

a) Using: { R }_{ series }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }....+{ R }_{ n }

{ R }_{ circuit }=10+12+14=36\Omega

b) Using V=IR and the total resistance for the circuit:

{ I }_{ circuit }=\cfrac { V }{ R }

{ I }_{ circuit }=\cfrac { 12 }{ 36 }

{ I }_{ circuit }=0.3A

c) Using V=IR

{ V }_{ 2 }=0.3\times 12

{ V }_{ 2 }=4V (using absolute value from part b)


Example 2: Parallel circuit:

a) Calculate the effective equivalent resistance of the circuit:

b) Calculate the total current flowing through the circuit:

c) What is the potential difference across { R }_{ 1 }:

d) What is the current through { R }_{ 2 }

Answers:

a) Using: \cfrac { 1 }{ { R }_{ parallel } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } ....+\cfrac { 1 }{ { R }_{ n } }

\cfrac { 1 }{ { R }_{ circuit } } =\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 12 }

\cfrac { 1 }{ { R }_{ circuit } } =\cfrac { 5 }{ 12 }

{ R }_{ circuit }=\cfrac { 12 }{ 5 } =2.4\Omega

b) Using V=IR and the total resistance for the circuit:

{ I }_{ circuit }=\cfrac { V }{ R }

{ I }_{ circuit }=\cfrac { 30 }{ 2.4 }

{ I }_{ circuit }=12.5A

c) As it is a simple parallel circuit, the potential difference across { R }_{ 1 } and { R }_{ 2 } will be the same and equal to the potential difference supplied to the circuit:

Therefore, the potential difference across { R }_{ 1 }=30V

d) Using V=IR

{ I }_{ 2 }=\cfrac { 30 }{ 12 }

{ I }_{ 2 }=2.5A


Example 3: Complex circuits (containing series and parallel sections)

a) Calculate the effective equivalent resistance of the circuit:

b) Calculate the total current flowing through the circuit:

c) What is the potential difference across each resistor?

d) What is the current through each resistor?

Answers:

a) First the equivalent effective resistance of the parallel section is determined and then added to the other series components:

\cfrac { 1 }{ { R }_{ 2\:and\:3 } } =\cfrac { 1 }{ 15 } +\cfrac { 1 }{ 10 }

\cfrac { 1 }{ { R }_{2\:and\:3 } } =\cfrac { 5 }{ 30 }

{ R }_{ 2\:and\:3 }=\cfrac { 30 }{ 5 } =6\Omega

{ R }_{ circuit }\:=4+6+20=30\Omega

b) Using V=IR and the total resistance for the circuit:

{ I }_{ circuit }=\cfrac { V }{ R }

{ I }_{ circuit }=\cfrac { 6 }{ 30 }

{ I }_{ circuit }=0.2A

c) Using V=IR and the resistance for each series component or parallel section:

{ V }_{ 1 }=0.2\times 4=0.8V

{ V }_{ 2\:and\:3 }=0.2\times 6=1.2V

{ V }_{ 4 }=0.2\times 20=4V

d) Using Ohm’s law: { I }_{ circuit }=\cfrac { V }{ R }

{ I }_{ 1 }={ I }_{ 4 }=0.2A

{ I }_{ 2 }=\cfrac { 1.2 }{ 15 } =0.08A

{ I }_{ 3 }=\cfrac { 1.2 }{ 10 } =0.12A

If we add { I }_{ 2 } and { I }_{ 3 } we get 0.08+0.12=0.2A which confirms Kirchoff’s current law.

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