Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell is

Options

A. 2.7 V
B. 2.1 V
C. 3V
D. 4.2 V

Answer

A

Explanation

Given that, the resultant voltage across the battery terminal = 1.5 V

Let l be the current in the circuit, then total resistance
`= 0.6 Omega`
Hence, `V= lR rArr 1.5=l xx0.6`
`l=(1.5)/(0.6)rArr l=(5)/(2)A`
Now, applying Kirchhoff's second law in the circuit.
`0.4l + 0.2l + 1.2-E=0`
`0.6xx(5)/(2)+1.2=E`
`rArr E = 2.7 V`