Electronics Engineering
Electrical Engineering

How do you calculate the total mA needed to power 3 9v DC 200 mA devices?

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January 30, 2011 5:43PM

Separate devices should always be hooked up in parallel, and not

series. Therefore, you would combine the total current requirement,

which in this case is 600mA or 0.6A. You would therefore need a 9V

power supply that is capable supplying at least 0.6A.

It depends on whether the devices are connected in series or

parallel or some combination of the two.

The current in a series circuit is the same throughout the

circuit. The current in a parallel circuit is the sum of the

current in each parallel branch

If the three devices are connected in series, the required

current is 200mA.

If the three devices are connected in parallel, the required

current is 600mA

The total power required by a circuit is the sum of the power

consumed by each component. It doesn't matter if the devices are

connected in series or parallel. Power = voltage x current.

If the three devices are in series, the power consumed by one of

them is 9 x .2 = 1.8 watts. The total power consumed is 1.8 x 3 =

5.4W

If the three devices are connected in parallel, the power

consumed is 9 x .6 = 5.4W

Notice that the power consumed is the same for the series and

parallel combination. The battery (or power supply) must be able to

supply 5.4 watts.


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