You are here:

Electronics/question regarding Ohms Law.

Advertisement


Question
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.  

Electronics

All Answers


Answers by Expert:


Ask Experts

Volunteer


ZZ

Expertise

Electronics questions about AC, DC and digital theory.

Experience

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.)

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

Education/Credentials
BSEE University of North Dakota

©2016 About.com. All rights reserved.