Question

N lamps each of resistance r, are fed by a machine of resistance 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, where n is the integer nearest to.

• A. ${\left(\frac{r}{NR}\right)}^{3/2}$
• B. ${\left(\frac{NR}{r}\right)}^{1/2}$
• C. $(NRr{)}^{3/2}$
• D. $(NRr{)}^{1/2}$

Answer 1

Answer: (B)

${\left(\frac{NR}{r}\right)}^{1/2}$

As each are containing n lamps, hence,
Resistance of each arc$=$nr
Number of arcs$=N/n$
Total resistance S is given by
$\frac{1}{S}=\sum \frac{1}{nr}=\frac{N}{n}\left(\frac{1}{nr}\right)$
$S=\frac{n^{2}r}{N}$.
$\therefore$ Total resistance $=R+S=R+\left(n^{2}r\right)N$
If E is the emf of the machine, current entering the arcs is $E/(R+S)$ and in each arc is $nE/(R+S)N$.
Hence, current passing through each lamp
$I=\frac{nE}{N(R+n^{2}r/N)}=\frac{E}{N}{\left[\frac{R}{n}+\frac{nr}{N}\right]}^{-1}$
Now, heat produced per second in the lamps is
$H=N_r{I}^2$
Since, light emitted is proportional to $H^2$ therefore light produced is maximum when $H^2$ and hence H is maximum or $\left[\frac{R}{n}+\frac{nr}{N}\right]$ is minimum. Hence, we can write,
$\frac{R}{n}+\frac{nr}{N}={[{\left(\frac{R}{n}\right)}^{1/2}-{\left(\frac{nr}{N}\right)}^{1/2}]}^2+2{\left(\frac{Rr}{N}\right)}^{1/2}$
This is minimum when $\sqrt{R/n}-\sqrt{nr/N}=0$
or very small or n is closely equal to $(NR/r{)}^{1/2}$.