# Question about changing resistance and power

Tags:
1. Apr 1, 2016

### MBBphys

1. The problem statement, all variables and given/known data
So this was the question:
The 60W filament lamp is connected to a 6.0 V battery. The resistance of the lamp in this circuit is 70 Ohms. Explain why this value differs from the value given in (a)(iv) when the lamp is connected to the 230V supply.

V=IR, P=IV, I=V/R
Therefore,
P=(V^2)/R
P is a constant here (60W)
Hence, (V^2) is directly proportional to R, so as we connect the lamp to a lower voltage battery, we measure a lower resistance.

However, the answer scheme said this:

The resistance of a metal increases with temperature. At 6V, lower heating effect, hence lower resistance.

So is there anything wrong with the argument I gave?

Thank you very much in advance!

2. Relevant equations
V=IR
P=IV

3. The attempt at a solution
(Shown above)

2. Apr 1, 2016

### Staff: Mentor

So the power dissipated by the bulb here is obviously not 60W, right? The bulb is probably not glowing when connected to only a 6V source...
What's (a)(iv)? Is there a figure that goes with this question?