Question

36 cells each of e.m.f. 2 volt and internal resistance 0.5 ohm are connected in mixed grouping so as to deliver maximum current to a bulb of resistance 2 ohm. How are they connected. What is the maximum current through the bulb (n=no. of bulbs in series and, m = no. of bulbs in parallel)

• A. n=12 ; m=3; 6 A
• B. n = 9; m=4; 5A
• C. n=5; m=7’ 6A
• D. n=7; m=4; 8A

Answer 1

The correct option is A n=12 ; m=3; 6 A

To deliver maximum current, the external resistance should be equal to internal resistance.
mn = 36; Rm = r n
$\frac{m}{n}=\frac{1}{4}=\frac{2}{8}=\frac{3}{12}\frac{m}{n}=\frac{r}{R}=\frac{0.5}{2}=\frac{1}{4}{l}_{max}=\frac{mnE}{2nr}=\frac{3\times 12\times 2}{2\times 12\times 0.5}6A\to \left(1\right)$