There are two electric bulbs of 40, W and 100 | vedclass

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(A) The resistance $R$ of a bulb is given by $R = \frac{V^2}{P}$. Since $V$ is constant,$R \propto \frac{1}{P}$.Thus,the resistance of the $40\, W$ bu

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$40\, W$ in series and $100\, W$ in parallel

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(A) The resistance $R$ of a bulb is given by $R = \frac{V^2}{P}$. Since $V$ is constant,$R \propto \frac{1}{P}$.Thus,the resistance of the $40\, W$ bulb $(R_1)$ is higher than the resistance of the $100\, W$ bulb $(R_2)$. Specifically,$\frac{R_1}{R_2} = \frac{100}{40} = 2.5$.In a series connection,the current $I$ is the same for both bulbs. The power dissipated is $P = I^2 R$. Since $R_1 > R_2$,the $40\, W$ bulb dissipates more power and glows brighter.In a parallel connection,the voltage $V$ is the same for both bulbs. The power dissipated is $P = \frac{V^2}{R}$. Since $R_2 Therefore,the $40\, W$ bulb is brighter in series,and the $100\, W$ bulb is brighter in parallel.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(P)$ $M^1 L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(Q)$ $M^1 L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(R)$ $M^1 L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(S)$ None of these

There are two electric bulbs of $40\, W$ and $100\, W$. Which one will be brighter when first connected in series and then in parallel?

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$(A)$ Electrical resistance $(P)$ $M^1 L^3 T^{-3} A^{-2}$
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$(C)$ Specific resistance $(R)$ $M^1 L^2 T^{-3} A^{-1}$
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