Question

In an experiment, N identical electrical bulbs, each having resistance R, are connected in parallel to a dc source of emf E and internal resistance r. If the power consumed by each bulb is $\frac{X{E}^{2}R}{(Nr+R{)}^2}$. Find out the value of X

Answer 1

Equivalent resistance of $N$ bulbs$=\frac{R}{N}$

Now,current in the circuit$=i=\frac{E}{\frac{R}{N}+r}$

$⟹i=\frac{NE}{R+Nr}$

This current gets distributed among $N$ equal resistance bulbs connected in parallel.

Hence, current through each bulb$=i_o=\frac{i}{N}=\frac{E}{R+Nr}$

Power consumed by each bulb$P_o=i_o^{2}R$

$⟹P_o=\frac{E^{2}R}{(R+Nr{)}^2}$

Hence, $X=1$