### Relating voltage, current and resistance

Setting up a simple circuit consists of selecting two components – the battery and the bulb, and linking them with wires. The two choices you make set the push

: the voltage of the battery, and how much that push is impeded: the resistance of the bulb. The current in the loop is the result of these two choices. Much more on this in the SPT: Electricity and energy topic.

The relationship between the push of the battery and the size of electric current is summarised by:

current = voltageresistance

You can write this out in symbols:

*I* = *V**R*

Where *V* is voltage, measured in volt; *I* is current, measured in ampere; *R* is resistance, measured in ohm.

According to this relationship, if the resistance does not change, then increasing the voltage results in an increase in current.

Choosing a 3 volt battery and a 10 ohm resistor results in a current of 0.3 ampere in the resistor (and in the battery, and in the connecting wires).

*I* = *V**R*

Put in the values to get:

current in loop = 3 volt10 ohm

and work out the current as 0.3 ampere. If the battery voltage is doubled to 6 volt, the new current is still given by*I*=

*V*

*R*, but you need to insert the new values current in loop now = 6 volt10 ohm and calulate the current as 0.6 ampere.

There is a 0.6 ampere current in each part of the circuit loop: in the lamp; in the battery; in both the connecting wires.

Double the battery voltage gives double the current.

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