Parallel resistance and heat problem

[KauGan]

Homework Statement

A heater joined in parallel with a 60W bulb is connected to the mains. If 60W bulb is replaced by a 100W bulb, will the rate of heat produced be more/less/remain the same ?


Potential Difference = 220V

Homework Equations

H= VIt = I^2Rt = V^2*t/R
P= VI

The Attempt at a Solution

R= (V^2/P)
= 48400/60
=806.6 ohms

R = V^2/P
= 48400/100
= 484 ohms

I=V/R
= 220/806.6
= 0.27 A
I = V/R
= 220/484
= 0.45 A

I am stuck here. I noticed that it is given 'rate of heat' which is rate of energy which is again nothing but power.
I don't know what I'm finding in the problem.
How do I calculate the rate of heat of the heater in parallel with the bulb ?
Has parallel connection of the bulb and heater anything to do with the problem, or is it just to confuse us?

 

[lewando]

You must first decide if the question is asking about the heat generated by the heater or by the total system (heater plus bulb).

 

[rl.bhat]

H = v^2*t/R.

In parallel combination, V is the same for all components. So by connecting 60 W bulb or 100W bulb will not change the resistance of the heater or the voltage across. So H remains the same.