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Electronics/question regarding Ohms Law.

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Hello Sir, I have already asked you this question but I am confused in your explanation that you gave last time. Please help me one more time in this question.
.
A 60 W bulb and a 100 W bulb are connected in series. Which of
them will glow more brightly when a constant pd is applied
across the circuit ?
.
No more values and clues are given in this question. What I did in solving this problem is I used a general formula: P = I^2 * R . In series, we have constant I for both bulbs, so P is directly proportional to R. So, 100 W bulb will have more R and hence glows more brightly. But, in your last email you told me that 60 W bulb will have high R. How is it possible ?

Answer
No, P is proportional to I**2 isn't it?  More R means it would have less current when connected to the same voltage - which is where the bulb rating comes from.  Remember, the rating of the bulb is computed with the standard voltage across it; not from that which it will have across it when in series with the other one!

E = I*r100 + I*r60

Just looking at the logic of the bulb ratings; if one bulb dissipates 100w then it must have a lower resistance than the 60w bulb.

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Electronics questions about AC, DC and digital theory.

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Graduate electrical engineer with over 40 years in electronic design, manufacturing, project organization and patent review. Experience in fields of industrial and consumer electronics (audio, video, acoustics, etc.)

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IEEE (Institute of Electrical and Electronics Engineers); Senior Life member AES (Audio Engineering Society), Fellow Life member

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BSEE University of North Dakota

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