Q108 / 160 · JIPMER 2004 · Physics

When a resistance of $9 \Omega$ is connected at the ends of a battery, its potential difference decreases from 40 V to 30 V. The internal resistance of the battery is

  • Correct answer
    A
    $3 \Omega$
  • B
    $6 \Omega$
  • C
    $1.5 \Omega$
  • D
    $4 \Omega$

Solution

$E=40 \mathrm{~V}, V=30 \mathrm{~V}, R=9 \Omega$ Internal resistance $r=\left(\frac{E}{V}-1\right) R=\left(\frac{40}{30}-1\right) 9$ $=\frac{1}{3} \times 9=3 \Omega$

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

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