Q106 / 160 · JIPMER 2013 · Physics

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

  • Correct answer
    A
    $1 \mathrm{~J} / \mathrm{s}$
  • B
    $5 \mathrm{~J} / \mathrm{s}$
  • C
    $0.42 \mathrm{~J} / \mathrm{s}$
  • D
    $0.8 \mathrm{~J} / \mathrm{s}$

Solution

Rate of energy $\frac{H}{t}=i^2 R$ (But total Resistance $R=21+4=25 \Omega$ ) So, $\quad \frac{H}{t}=0.2 \times 0.2 \times 25=1 \mathrm{~J} / \mathrm{s}$

Hence, the correct answer is option: 'A'.

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