JEE MainPhysicsModerateCurrent Electricity

The emf of cell E_2 is

Show answer and solution

i = \frac{2}{15 + 10}

V_{AB} = i10 = \frac{20}{25}

E_{2} = \frac{V_{AB}}{50} \times 31.25

= \frac{20}{25} \times \frac{31.25}{50}

=0.5V

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

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.