Calculating Current in a Series Circuit with Different Bulb Resistances

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SUMMARY

The discussion focuses on calculating the current in a series circuit containing a 60W-120V bulb and a 200W-120V bulb connected to a 240V supply. The correct approach involves determining the resistance of each bulb using the formula R = V^2 / P, leading to resistances of 240 ohms for the 60W bulb and 144 ohms for the 200W bulb. The total resistance in the circuit is 384 ohms, resulting in a current of 0.625A, which is confirmed through Ohm's Law (I = V / R). Misunderstandings about voltage distribution and power calculations were clarified during the discussion.

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Romperstomper
A 60W-120V bulb and a 200W-120V bulb are connected in a series across a 240V line. Assume that the resistance of each bulb does not vary with current.(Note:The 60W-120V means it dissipates 60W when connected to a 120V line, and the 200W-120V means it dissipates 200W when connected to a 120V line). Find the current running through the bulbs.

What I did: The voltage is double, so I figured that if the voltage is doubled, the power must be doubled since [tex]P = I * \Delta V[/tex] and the current remains the same. Since [tex]\Delta V[/tex] changes between each bulb, I decided to use the net power and solved for I.

[tex]520 = I * 240 = 2.17A[/tex]. This is way off from the answer of .769A. Can anyone tell me what I did wrong?
 
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Hooking up light bulbs in series is equivalent to hooking up to resistors in series - i.e. the total resistance is the sum of the individual resistances.

Also, the total voltage drop across both lights will be 240 V - NOT double! I.e. the sum of the potential drops the two bulbs must add up to 240 V.

Use the data provided to determine the resistance of each light bulb, add them together and use Ohm's Law to find the current through the circuit.
 
Ok, I see what I was doing wrong. I reworked it out and got the right answer. Thanks for the help.
 

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