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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Current Electricity

JIPMER 2013 · Physics · Current Electricity

The internal resistance of a primary cell is 4 . It generates a...

The internal resistance of a primary cell is 4Ω4 \Omega. It generates a current of 0.2 A in an external resistance of 21Ω21 \Omega. The rate at which chemical energy to consumed in providing current is

Options

  1. A1 J/s1 \mathrm{~J} / \mathrm{s}
  2. B5 J/s5 \mathrm{~J} / \mathrm{s}
  3. C0.42 J/s0.42 \mathrm{~J} / \mathrm{s}
  4. D0.8 J/s0.8 \mathrm{~J} / \mathrm{s}

Correct answer

A. 1 J/s1 \mathrm{~J} / \mathrm{s}

Step-by-step solution

Rate of energy Ht=i2R\frac{H}{t}=i^2 R (But total Resistance R=21+4=25ΩR=21+4=25 \Omega ) So, Ht=0.2×0.2×25=1 J/s\quad \frac{H}{t}=0.2 \times 0.2 \times 25=1 \mathrm{~J} / \mathrm{s}
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