Circuit Question

Could someone tell me the relationship between bulb resistance and the brightness of a bulb in a parallel, series and combination circuit?

Reply 1

Stonebridge

Original post by maidoftheseas

Could someone tell me the relationship between bulb resistance and the brightness of a bulb in a parallel, series and combination circuit?

Welcome to TSR physics.

I think the question is a little too vague to produce any meaningful or helpful answer.
The best thing to do is ask a more specific question - perhaps post a question from your book or an exam that asks about this topic, and then say what it is you don't understand.

Reply 2

Fysiika07

brightness of bulb is related to power dissipated at bulb, power dissipated at bulb is related to resistance of bulb in two ways, P=I2R and P=V2R. In series connection, use P=I2R, as current is the same, power directly proportional to resistance. In parallel circuit, use P=V2R, power inversely proportional to resistance, since pd is the same.

Reply 3

maidoftheseas

Original post by Stonebridge

HI, yep I've got an example question. I've pasted the link below and the question is Unit 8, Question 27 (page 27). A explanation would be a lot of help :biggrin:

http://papers.xtremepapers.com/GAMSAT/ACER%20GAMSAT%20Sample%20Questions.pdf

Reply 4

Physicsaholic

Original post by Fysiika07

Just wanted to point out a small typographical error you made. :smile:

P=I^2R = V^2/R

Reply 5

Stonebridge

Well, any explanation needs to be in terms of what you have already studied, and this exam paper is one I've never seen before from a course I'm not familiar with. You need to tell us if you have already studied the rules for combining resistances in series and parallel and if you are able to apply Ohm's Law to work out current from V and R.
The brightest bulb in this question will be the one with the greatest pd across it or the largest current in it. It should be possible to almost guess this from inspection of the diagram. If not just make up some values for the resistance and cell voltage and see what happens.

Reply 6

uberteknik

Original post by maidoftheseas

HI, yep I've got an example question. I've pasted the link below and the question is Unit 8, Question 27 (page 27). A explanation would be a lot of help :biggrin:

[URL="http://papers.xtremepapers.com/GAMSAT/ACER%20GAMSAT%20Sample%20Questions.pdf

http://papers.xtremepapers.com/GAMSAT/ACER%20GAMSAT%20Sample%20Questions.pdf

CASE I: Both globes are in parallel and both connected across the supply therefore V_G = V_S Since both globes have the same voltage potential across them, the current through both and the power developed in both will also be the same.

Power = V_G²/R. and since V_G and R are the same values for both globes, they will glow with equal brightness.

CASE II: Both globes are in series, and since the resistances are identical, the pd developed across each globe will be V_G = V_s / 2

Power = (V_S/2)²/R and hence both globes will be quarter the brightness of CASE I.

CASE III and IV: Kirchoffs voltage law says the supply voltage must equal the sum of the voltage drops around the circuit. By inspection, the voltage pd developed across all three globes will be less than the supply and in the ratio 2V_S/3 for globe III and V_S/3 for globe IV.

Hence since power = V_G²/R and V_G < V_S in all three cases, then all three globes will be less bright than CASE I.

CASE I is the brightest globe.

(edited 9 years ago)

Reply 7

Fysiika07