## Resistors in parallel

##### Posted on : Sat , 01 2014 by : virusi

## Theory of operation :

When designing we very often use resistors therefore it is very useful to have a set of rules for finding the equivalent resistance. The circuit below shows the parallel connection of the resistors and the direction of the current.

Current splits as it travel through each breach so the sum of the currents through the branches is the same as the current at the input :
Also at any point the voltage through each resistor in parallel circuit must have the same voltage :
Because at any point the voltage through each resistor in parallel is the same we can remove the voltage from the equation :
Equation (4) can be generalized :
The equation (5) can be overwritten:
Resistance is the reciprocal of conductance so we can rewrite the equation (5):

Fig .1 Resistors in parallel

Fig .2 Another connection of resistors in parallel

Resistors can be connected such that they branch out from a single point (known as a node), and join up again somewhere else in the circuit as you can see from the fig.1 and fig.2. Current splits as it travel through each breach so the sum of the currents through the branches is the same as the current at the input :

\large I=I_{1}+I_{2}+I_{3}+...+I_{n}(1)

\large V_{1} = V_{2} = V_{3} = ... = V_{n}(2)

By Ohm’s Law (

R_{equivalent} = \frac{V}{I}) equation 2 can be overwritten :

\large \frac{V}{R_{equivalent}} = \frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}+...+\frac{V_{n}}{R_{n}}(3)

\large \frac{1}{R_{equivalent}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}+...+\frac{1}{R_{n}}(4)

\frac{1}{R_{equivalent}} = \sum \frac{1}{R_{n}}(5)

\frac{1}{R_{equivalent}} = \sum \frac{1}{\frac{1}{R_{n}}}(6)

\large g_{equivalent} = \sum g_{n}(7)

**Tips and tricks :** – A larger resistor in parallel with a smaller resistance has the resistance of the smaller one, roughly.

– If you have a 10k in parallel with a 20k resistance. You can consider the 10k resistance like two 20k’s in parallel than we have 3 resistances of 20k in parallel. Because the resistance of n equal resistors in parallel is :
So in our case the total resistance of the circuit is equal to 20k/3 resistor or a 6.6k resistor.

\large R = \frac{1}{\frac{n}{R_{1}}} = \frac{R_{1}}{n}(8)

## Practical Example :

If you don’t see the example below than you should follow this steps:
You can see 2 circuits but both of them as we already know are equivalent. Why? as you can see the current and the voltage that flows through both circuits are the same. 1 diagram is showing the current and voltage that is flowing through first circuit. 2 diagram is showing the current and voltage that is flowing through second circuit. If you want to change any resistance value or the input voltage than just double click on the object and insert the desired value.

– In your browser allow Java SE 7.

– Lower you java security settings (Go to Control Panel >> Java >> Security and set the security level to medium) .

– Edit Site List (Go to Control Panel >> Java >> Security and click on Edit Site List… and add eagerlearning.org in the list).

Last updated on Mon , 03 2014

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