Parallel Resistance for Two Identical Headlights with 6V Battery

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SUMMARY

The discussion focuses on calculating the resistance of two identical headlights connected in parallel to a 6V battery supplying a total power of 48W. The power formula P = VI is applied, leading to a total current of 8A, which results in 4A flowing through each headlight. Consequently, the resistance of each bulb can be determined using Ohm's Law, confirming that each headlight has a resistance of 1.5 ohms.

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nautica
A 6 volt battery supplies a total of 48W to two identicle headlights in parrallel. The resistance in each bulb is:

My confusion here is that I am not sure where the watts would fit into resistances.

Could someone lead me in the right directions.

thanks
nautica
 
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Use the power to find the current through each lightbulb (P = VI); then use Ohm's law to find resistance.
 
So

48W=6VI, so 8 total, which would give 4 in each, right?

thanks
nautica
 

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