Two cells `epsi_1` and `epsi_2` in the given circuit diagram have an emf of 5 V and 9 V and internal resistance of 0.3 `Omega` and 1.2`Omega` respectively. Calculate the value of current flowing through the resistance of 3 `Omega` .

Explanation

Here net emf of circuit `epsi = epsi_2 - epsi_1 = 9 - 5 = 4 V` and total resistance of the circuit
`R = (6 xx 3)/(6 + 3) + 4.5 + 0.3 + 1.2 = 8Omega`
` therefore ` Main circuit current `I = epsi/R = (4V)/(8Omega) = 0.5A`
If current flowing through `3Omega` resistance be `I_1` , then current flowing through `6Omega` resistance will be `(0.5 - I_1)` and hence
`3I_1 = 6 xx (0.5 - I_1) rArr I_1 = 0.33 A`