$10$ identical cells are connected in series with similar polarity to an external resistance. Current through this resistance is $^{'} I^{'}$ . Now polarity of two cells are reversed, calculate current through same resistance.

0.2 I

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0.4 I

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0.6 I

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0.8 I

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Solution

The correct option is C $0.6 I$
With same polarity,
$I = \frac{Net emf}{Total resistance} = \frac{10 E}{R + 10 r} . . (i)$

When polarity of two cells are reversed,
$I^{'} = \frac{8 E - 2 E}{R + 10 r} = \frac{6 E}{R + 10 r} . . (i i)$
From equation $(i)$ and $(i i)$
$\Rightarrow I^{'} = 0.6 I$

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10 identical cells are connected in series with similar polarity to an external resistance. Current through this resistance is ′I′. Now polarity of two cells are reversed, calculate current through same resistance.

The correct option is C 0.6IWith same polarity, I=Net emfTotal resistance=10ER+10r ..(i) When polarity of two cells are reversed, I′=8E−2ER+10r=6ER+10r ..(ii) From equation (i) and (ii) ⇒I′=0.6I

$10$ identical cells are connected in series with similar polarity to an external resistance. Current through this resistance is $^{'} I^{'}$ . Now polarity of two cells are reversed, calculate current through same resistance.

The correct option is C $0.6 I$
With same polarity,
$I = \frac{Net emf}{Total resistance} = \frac{10 E}{R + 10 r} . . (i)$

When polarity of two cells are reversed,
$I^{'} = \frac{8 E - 2 E}{R + 10 r} = \frac{6 E}{R + 10 r} . . (i i)$
From equation $(i)$ and $(i i)$
$\Rightarrow I^{'} = 0.6 I$