A cell $$E_1$$ of emf 6 V and internal resistance 2$$\Omega$$ is connected with another cell $$E_2$$ of emf 4 V and internal resistance 8 $$\Omega$$ (as shown in the figure). The potential difference across points X and Y is:

From the figure, both cells are connected in opposition (their positive terminals face each other at point X).

step 1: net emf

E₁ = 6 V, E₂ = 4 V (opposing)

Net emf = 6 − 4 = 2 V

step 2: total internal resistance

r_total = 2 + 8 = 10 Ω

step 3: circuit current

I = net emf / total resistance = 2 / 10 = 0.2 A

Direction: from stronger cell (6 V) toward weaker cell

step 4: find potential difference between X and Y

Move from X → Y across E₂ (4 V, internal resistance 8Ω)

Across E₂:
• potential drop due to emf = 4 V (since we go from + to −)
• additional drop due to internal resistance = I × 8 = 0.2 × 8 = 1.6 V

Total drop = 4 + 1.6 = 5.6 V