JEE MainPhysicsModerateCurrent Electricity

The internal resistance of cell E2

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[IMAGE] V_AB = 2

i^{'} = \frac{0.5}{r + 5}

0.5 - i^{'}r

= \frac{V_{AB}}{50} \times 5

0.5 - \frac{0.5}{r + 5}r = \frac{2}{10}

0.5 - 0.2 = \frac{0.5r}{r + 5}

0.3 r +1.5=0.5 r
0.2r = 1.5
r = 7.5

\Omega

Source: Grass-G / P-12 Physics paper 2 Nov.docx

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