# GCSE - Electricity question

Hi,

I was doing a past GCSE paper and I got stuck on this question to do with electricity, so I was wondering if anyone would be able to help.

'When the voltage across the bulb falls to half, the current through the bulb falls by less than half. Why is this?'

Thanks!
u got me stumped and i do A Level fysix...lol
Lol, same.
I dunno, maybe it's like a more practical reason. Such as resistance across the bulb increasing, as it heats up. Since I = V/R, a higher R will yield a smaller I...
lol, same here too.

Err... V=IR

Do u think this makes sense.

if the voltage is reduced, it is doing less work against the charge particles flowing in the circuit. because the harder the voltage is the greater the internal energy which will further oppose the movement of charge particles. reducing the voltage will mean that there is less internal energy(heat) created therefore the current wont decrease as much.

Correct anyone??????
mockel
Lol, same.
I dunno, maybe it's like a more practical reason. Such as resistance across the bulb increasing, as it heats up. Since I = V/R, a higher R will yield a smaller I...

hey mockel, just read urs and its kinda the same as mine, lol
andyj72
lol, same here too.

Err... V=IR

Do u think this makes sense.

if the voltage is reduced, it is doing less work against the charge particles flowing in the circuit. because the harder the voltage is the greater the internal energy which will further oppose the movement of charge particles. reducing the voltage will mean that there is less internal energy(heat) created therefore the current wont decrease as much.

Correct anyone??????

Thats right, but waaaaaay to much depth for GCSE.
SinghFello
Thats right, but waaaaaay to much depth for GCSE.

lol, yes. i will try to explain in in brief.

u can c my luv of phy 2 - electricity and thermal phy
sweet_gurl, i will try to simplify the explantion.

Basically,

if the voltage is reduced, it is doing less work against the charge particles flowing in the circuit. this is because the higher the voltage, the heat build up is more , more particles are moving towards the resistance heating it up. so reducing the voltage means that the components will not heat up as much.(increase in heat reduces the flow of charge). so more current will not decrease as much. .

hope this helps. if u can understand this u can c if u can make sense of what i said previously. the examiner will be impressed and give u full marks for the question and even a bonus mark