Assertion : An electric bulb becomes dim,when | vedclass

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(A) The electric power of a heater is significantly higher than that of a bulb. Since $P = V^2/R$,we have $R \propto 1/P$,meaning the resistance of th

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(A) The electric power of a heater is significantly higher than that of a bulb. Since $P = V^2/R$,we have $R \propto 1/P$,meaning the resistance of the heater is much lower than that of the bulb.When the heater is switched on in parallel,the total current drawn from the source increases. Due to the internal resistance of the source (or line resistance),the terminal voltage across the bulb drops,causing it to become dim.As the heater coil heats up,its resistance increases due to the positive temperature coefficient of the material. Consequently,the current drawn by the heater decreases,the terminal voltage recovers,and the bulb's brightness increases (dimness decreases).

List-$I$	List-$II$
$(I)$ Two wires of same resistance are connected in series and same current is passed through them	$(A)$ $1:2$
$(II)$ Two wires of resistance $R$ and $2R$ $\Omega$ are connected in series and same $P.D.$ is applied across them	$(B)$ $4:1$
$(III)$ Two wires of same resistance are connected in parallel and same current is flowing through them	$(C)$ $1:1$
$(IV)$ Two wires of resistances in the ratio $1:2$ are connected in parallel and same $P.D.$ is applied across them	$(D)$ $2:1$

Assertion : An electric bulb becomes dim,when the electric heater in parallel circuit is switched on.
Reason : Dimness decreases after sometime.

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Assertion : An electric bulb becomes dim,when the electric heater in parallel circuit is switched on.Reason : Dimness decreases after sometime.