36 cells each of e.m.f. 2 volt and internal resistance 0.5 ohm are connected in

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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)

Options:

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: Option A
:
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_{m a x} = \frac{m n E}{2 n r} = \frac{3 \times 12 \times 2}{2 \times 12 \times 0.5} 6 A \to (1)

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