In the network shown in the above figure find the current through ` 3 Omega` and `10 Omega` resistances.

Explanation

The network is a simple and parallel combination of resistors.
Total resistance ` = [((10 xx 15)/(10+ 15)) + 3 + 1] omega = (6+4) Omega`

Current I from A to C , `I = (10 V)/(10 Omega ) = 1A` (Current through `3 Omega`)
Potential difference across `V_(AB) = (1 A)(6 Omega) = 6V`
`10 Omega ` and `15 Omega` are in parllel across AB. So, potential difference across both of then ` = V_(AB) = 6 V` .
Current through `10 Omega , l = (V_(AB))/(10 Omega) = (6 V)/(10 Omega) = 0.6 A`