Two cells `E_(1)` and `E_(2)` in the circuit shown in figure, have emfs of 5 V and 9 V and internal resistance of `0.3 Omega` and `1.2 Omega` respestivley. Calculate the value of current flowing through the resistance of `3 Omega`.

Answer

` (1)/(3) A`

Explanation

As the two cell are connected in oppostion Net emf `= E_(2) - E_(1) = 9 - 5 = 4V`
Total resistance `= 0.3 + 1.2 + 4.5 + (6 xx 3)/(6 + 3) = 8 Omega`
Current through circuit, `I = (4)/(8) = 0.5 A`
Current through circuit `3 Omega` resistance `= (6 xx 0.5)/(6 + 3) = (1)/(3) A`