- #1
KauGan
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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?