The circuit diagram has two cells `epsilon_(1) and epsilon_(2)` with emf `4 V and 2V` respectively, each one having an internal resistance `2 Omega `. The external resistance `R `is of `8 Omega`. Find the magnitude and the direction of current flowing through the two cells

Answer

`l_(1) = (2)/(3) A`, `l_(2) = (1)/(3) A`

Explanation

Here `epsilon_(1) = 4 V, r_(1) = 2 Omega , epsilon_(2) = 2V, r_(2) = 2 Omega, R = 8 Omega`
Using kichoff's law of loop on the loop `ABCFA` we have
or `- epsilon_(1) + I_(1) r_(1) + epsilon_(2) + I_(2) r_(2) = 0`
` or - 4+ I_(1)xx 2 + 2 - I_(2) xx 2 = 0`
or `2I_(1) - 2I_(2) = 4 - 2 = 2`
or `I_(1) - I_(2) = 1`......(i)
For loop `CDEFC`
`- epsilon_(2) + I_(2) r_(2) + (I_(1)- I_(2)) R = 0`
or `- 2+ I_(2) xx 2 +( I_(1) + I_(2)) 8 = 0`
or `8I_(1) + 10I_(2) = 2 or 4I_(1) +5 I_(2) = 1` ......(ii)
On solving (i) and (ii) we get
`I_(1) = (2)/(3) A` (from negative to positive terminal insides cell `epsilon_(2)`)
`l_(2) = (1)/(3) A` (from `(+v)` to `(-v)` terminal insides cell `epsilon_(2)`)