Three cells A, B and C of emfs 2 V, 3 V and 5 V respectively are connected in parallel to each other. Their internal resistances are 5 Ω, 5 Ω and 1 Ω…

In loop ABEFA,\[5 - 3 - 5I_1 - I = 0\] \[2 = 5I_1 + I \qquad ----- (1)\]In loop CBEDC,\[2 - 3 - 5I_1 + 5I - 5I_1 = 0\] \[-1 = 10I_1 - 5I \qquad ----- (2)\]Solving equation ($\displaystyle 1$) and ($\displaystyle 2$)$\displaystyle I = \dfrac{5}{7}\,\text{A}$ in arm AF/through the cell of 5V(A)$\displaystyle I_1 = \dfrac{9}{35}\,\text{A}$ in arm BE/through the cell of 3V(B)$\displaystyle I - I_1 = \dfrac{16}{35}\,\text{A}$ in arm CD/through the cell of 2V(C)