# Watts triangle

Discussion in 'General Electronics Chat' started by Webby, Sep 6, 2008.

1. ### Webby Thread Starter Active Member

Jun 15, 2008
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0
Hi all, with a 12v circuit with two 85 watt bulbs in parralel do I add the bulbs values together before calculating the current flowing or just divide by one bulb value only? Trying to work out the current flowing to use a suitable fuse rating.

Power divided Volts = current

85/12= 7.08amps

or

170/12=14.1amps

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214

Is your 12V source capable of delivering that much current?

hgmjr

3. ### Webby Thread Starter Active Member

Jun 15, 2008
66
0
Hi, see what you mean this question was on a test paper I had to compleat thats all what fuse rating would be suitable?

If this was a series circuit i presume i use the same equation adding the bulb wattage together?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If the bulbs were is series you would not use the same equation. Current in both bulbs would be the same but the voltage would be cut in half.

hgmjr

5. ### Webby Thread Starter Active Member

Jun 15, 2008
66
0
Thanks if in the paralel circuit we have 14.1amps currnt flow what fuse would be suitable 15amps or less?

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
A fuse less than 15 amps might prove too small for the parallel case. There are no absolute standards for fuse selection but typically you are safe to use a fuse large enough to insure that the heat rise in the interconnecting wires remains below 10 to 20 degrees celcius above ambient in the event of a short circuit.

hgmjr

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The fuse manufacturer Littelfuse has a pretty good Whitepaper on the subject of fuses at this link.

hgmjr

8. ### iamspook Member

Aug 6, 2008
27
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Assuming this is (say) a tungsten globe, then there are two characteristics to consider.
Steady-state - which is governed by the resistance once the bulb has reached steady temperature, and the non-linear surge that happens when you first apply voltage. For steady state - consider the resistances as lumped components. For considering surge, then
take the peak current drawn just after voltage is applied. The resistance of the filament falls as it heats up.

9. ### Webby Thread Starter Active Member

Jun 15, 2008
66
0
Hi, was trying to find the answer in the "eebook of all about circuits" why would you have to half the system voltage to work out the current flow if two bulbs are in series.

Do I add the bulb power i,e 85watts togther =170 watts divide by 6v= 28.3amps cuurent flow.