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Minibrute power question


andrew ward

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This is embarrasing, i should know this....

 

I just found out i have a 4 light and 6 light minibrute. Ie 6x650w bubbles. It has a 15amp lead for all the power and i was baffled as to how it ran (im in Australia (240v) but changing a bubble ive realised its 120 bubbles wired in series.

 

That means instead of drawing 3900w at 240v with all 6 bulbs on, it draws 1950w. Is that correct?

And if so, why does it have a 15amp lead as opposed to a 10amp lead (10amp being 2400w in 240v).

 

Have i misfigured the 1950w?

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Ok i think i've understood
the 650 bubbles maybe are Dwe par36 120v like you already know.
In couples of two because you like me use 240v ok
4 bubbles 2600w 6 bubbles 3900w if you use in 220v like in 120v the difference is in the ampere at 120v you need double ampere than 240v and this is the answer of the plug question.
So you have the same amount of watt and footcandle but you use double ampere than plug 6 650w at 240v.
I hope you understand my strange english.

Leonardo Brocato
Gaffer Rome Italy

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In a parallel circuit the voltage through each component of the circuit is equal to the total voltage of the system, while the total current is equal to the sum of the current draw of the individual components.

 

In a series circuit the current through each component of the circuit is equal to the total current of the system, while the total voltage is equal to the sum of the voltages.

 

So in the case of a DWE globe it draws 5.4A at 120V so in series the circuit total of the two globes would be 5.4A at 240V.

 

Now in a mini brute you have 3 of these circuits in parallel, so now the voltage remains constant and the current is added together. So for the lamphead you get to a total of 16.2A at 240V.

 

The total power(wattage) of the system, whether in series or parallel is the total of the power of all the components. So for the mini brute it is 6 x 650w, so 3900w.

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