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The internal resistance of a primary cell is $4 \Omega$. It generates a current of 0.2 A in an external resistance of $21 \Omega$. The rate at which chemical energy to consumed in providing current is
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The correct answer is:
$1 \mathrm{~J} / \mathrm{s}$
Rate of energy $\frac{H}{t}=i^2 R$
(But total Resistance $R=21+4=25 \Omega$ )
So, $\quad \frac{H}{t}=0.2 \times 0.2 \times 25=1 \mathrm{~J} / \mathrm{s}$
(But total Resistance $R=21+4=25 \Omega$ )
So, $\quad \frac{H}{t}=0.2 \times 0.2 \times 25=1 \mathrm{~J} / \mathrm{s}$
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