 Home Page News Search Contact Us      Electrical riddle no. 8 - Lamp glow

Which will glow brighter and why?show the calculation involved?  #1
Sun, November 25, 2007 - 16:37
The lamp 60W will be brighter than 100W; because:

R60W > R100W  &  (P60W = R60W.I2 ) > (P100W = R100W.I2) But let us assume the lamp 60w is installed in room A and another lamp is located in room B with same electrical circuit. For maximum lighting achievement in room B, which lamp (rated power) shall be permuted with lamp 100w?

#2
Sun, January 13, 2008 - 19:33
Replace the 100W lamp  with a 60 W lamp
#3
Sun, February 10, 2008 - 07:26
i have done this before and the 100 watt bulb will burn out with 240 v breaking the circuit befor burning out the 60 watt
#4
Mon, March 24, 2008 - 02:35
(a)The question is rather misleading. It can be interpreted in two different scenarios:

(i) A 240V circuit supplying two lamps expending 60W and 100W power respectively. In this case, the 100W lamp will be brighter with a resistance =11.875Ω, 60W with R=7.125Ω.

(ii) Two 240V lamps rated at 60W and 100W respectively are connected in series. In this case, the result will be as Hamid has given. Both rooms will have a 15W lighting outpu, total for two rooms is 30W.

(b) Answer to Hamid's question, based on (ii) above, use maximum power transfer principle, assume the source is an ideal source, add a 150W in series with 100W in room B.

(c) Bill, the bulbs will not burn out unless they were at a much lower voltage rating (i.e., 50V rated). Since they are connected in series, they cannot burn out separately. They will need to be burnt out simultaneously.

#5
Sat, September 6, 2008 - 07:05
first of all, given is so vague that complete rating of the bulb was not mentioned. it is essential that voltage rating ( or current) of the bulb be clearly stated. for the sake of discussion, let us say that each bulb are rated 60W and 100W and 240V each..

calculating the rated resistance of each lamp using P = V2/R, we can calculate the resistance of the 60W lamp as 960 ohms and resistance of the 100W lamp as 576 ohms.

so when they are connected in series across a 240V source, total resistance would be equal to 1536 ohms giving a current ( I = V/R) of 0.1563 ampere.

getting again the power consumed by each lamp when connected in series across a 240V source using I2R formula, the power across 60W lamp will now be 23.42 watts only while the power on the 100W lamp is 14.05 watts only.

considering the setup, the 60W lamp will glow brighter than the 100W lamp.

it would arrive into a different answer when configurations where known and altered like:

1. bulbs are rated 120V only

2. 1st bulb is 120V and another bulb is 240V

3. or author is thinking of a given like 60Ω and 100Ω rather than 60W and 100W  where in this case the 100Ω bulb will consume 225Watts while 60Ω is 135 watts only, thus 100Ω is brighter.

-------------------------
YEBAH..... #6
Wed, June 22nd, 2016 - 18:21
Light bulbs are non-linear devices - as they warm up so their resistance increases.  This, now, creates a whole new viewpoint for the initial question.  First one has to establish the temperature - resistance relationship of each light bulb (tungsten filament?) and then the temperature - current relationship of each bulb to get a current - resistance relationship for each bulb.
Now which one's brighter?  That's a bit more challenging!! #7
Thu, June 23rd, 2016 - 11:21
Yes, it could be a new theoretical challenge! but the correct answer is same as I mentioned later. In fact, the rate of resistance variation due to temperature is so slow in comparison with 60 w and 100 w lamps resistance difference.    