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(A) The power rating of a bulb is given by the formula $P = V^2 / R$,where $V$ is the voltage and $R$ is the resistance of the filament.Assuming both

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$100\ W$ bulb has a thicker filament

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(A) The power rating of a bulb is given by the formula $P = V^2 / R$,where $V$ is the voltage and $R$ is the resistance of the filament.Assuming both bulbs are designed for the same operating voltage $V$,we have $R = V^2 / P$.This shows that resistance $R$ is inversely proportional to power $P$ $(R \propto 1/P)$.Therefore,the $100\ W$ bulb has lower resistance compared to the $60\ W$ bulb.The resistance of a wire is given by $R = \rho \ell / A$,where $\rho$ is the resistivity,$\ell$ is the length,and $A$ is the cross-sectional area (thickness).Since the length $\ell$ and material (resistivity $\rho$) are the same for both,$R \propto 1/A$.Since the $100\ W$ bulb has lower resistance,it must have a larger cross-sectional area $A$,meaning it has a thicker filament.

Two electric bulbs have tungsten filaments of the same length. If one of them is rated $60\ W$ and the other $100\ W$,then:

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