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 volt
B. 2.1 volt
C. 3 volt
D. 4.2 volt

Answer

B

Explanation

`E_(eq) = (E_(1)/r_(1) + E_(2)/r_(2))/(1/r_(1) + 1/r_(2))`
`rArr 1.5 = (r_(2)E_(1) + E_(2)r_(1))/(r_(1) + r_(2))`
`rArr 1.5 = (0.4 xx 1.2 + E_(2) xx 0.2)/0.6 rArr 0.9 = 0.48 + E_(2) xx 0.2`
`rArr E_(2) = (0.9 - 0.48)/0.2 = 2.1` Volt