Two cells of 6 V and 4 V having internal resistance of 3`Omega` and 2`Omega` respectively are connected in parallel so as to send a current through an external resistance 8`Omega` in the same direction. Find the current through the cells and the current through the external resistance.

Explanation

From Kirchoff.s loop rule,
To the loop ABEFA, we get
`(I_1 + I_2) 8 + 3I_1 = 6`
`8I_1 + 8I_2 + 3I_1 = 6`
`11I_1 + 8I_2 = 6` ……(1)
The loop BEDCB, we get
`(I_1 + I_2) 8 + 2I_1 = 4`
`8I_1 + 8I_2 + 2I_2 = 4`
`8I_1 + 10I_2 = 4` .....(2)

`I_1 = 36/44`
`I_1 = 0.818 A`
From (2)` 8(0.818) + 10I_2 = 4`
`6.544 + 10I_2 = 4`
`10I_2 = 4 - 6.544`
`10I_2 = -2.544`
`I_2 = -0.2544A`
The current through the external resistance
`I = I_1 + I_2 = 0.818 - 0.2544`
`I = 0.5636 A`