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10 identical cell in series are connected to the ends of a resistance of 59\Omega,the current is  found to be 0.25A. But when the same cells being connected in parallel,are joined to the ends of a resistance of 0.05\Omega,the current is 25A EMF of  each cell is

A.

1V

B.

0.5V

C.

1.5V

D.

2V

Answers (1)

 In series: I=\frac{E_{eq}}{R+r_{eq}}

E_{eq}=10E \: \: and\: \: r_{eq}=10r,I=0.25A

\therefore 0.25=\frac{10E}{59+10r}\: \: or\: \: 10E=2.5r+14.75.....(1)

In parallel: E_{eq}=E \: \: and\: \: r_{eq}=r/10,I=25A

\therefore 25=\frac{E}{0.05+\frac{r}{10}}\: \: or\: \: E=1.25+2.5r.....(2)

(1)-(2)

13.5=9E\: \: or\: \: E=1.5V

Posted by

Safeer PP

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