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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
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B
n = 9; m=4; 5A
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C
n=5; m=7’ 6A
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D
n=7; m=4; 8A
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Solution

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

mn=14=28=312mn=rR=0.52=14lmax=mnE2nr=3×12×22×12×0.56A(1)

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