N lamps each of resistacne r, are fed by a machine of resistacne R. If light emitted by any lamp is proportional to the square of the heat produced, p

Question

N lamps each of resistacne r, are fed by a machine of resistacne R. If light emitted by any lamp is proportional to the square of the heat produced, p

N lamps each of resistacne r, are fed by a machine of resistacne R. If light emitted by any lamp is proportional to the square of the heat produced, prove that the most efficient way of arranging them is to place them in parallel arcs, each containing n lamps, wheren is the integer nearest to
A. `(r)/(NR)^(3//2)`
B. `(NR/r)^(1//2)`
C. `(NRr)^(3//2)`
D. `(NRr)^(1//2)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - B
As each are containing n lamps , hence Resisitance of each arec=nr.Number of `arcs=N//n` Total resisitacne s is given by
`(1)/(s)=sigma(1)/(nr)=(N)/(n)((1)/(nr))rArrS=(n^(2)r)/(N)`
`therefore Total resistance =r+S=R+(n^(2)r)N`
If E is the emf of the machine current entering the arcs is `(E//R+S)` and in each arc is `(E//R+S)N`. Hence ,current passing through each lamp
`I=(nE)/(N(r+n^(2))(r//N)=(E)/(N)[(R)/(n+(nr))/(N)]^(-1)`
Now heat produced per second in the lamps is `H=N,I^(2)`
Since, light emitted is proportional of `H^(2)` therefore lighty produced is maximum when `h^(2)` and hence H is maximum or `[(R)/(n)+(nr)/(N)]` is minimum.Hence we can write
`R/n+(nr)/(N)=[((R)/(n))^(1)/(2)-((nr)/(N))^(1/2)]^(-2)+2((R)/(n))+2((Rr)/(N))^(1)/(2)`
This is minimum when `sqrt(R//n)-sqrt(nr//N)=0`
or very small or very small or n is closely equal to `(NR//R)^(3)/(2)`