Two cells of e.m.f. `E_(1)` and `E_(2), E_(1) gt E_(2)` having an internal resistance of `1Omega` each form a closed circuit with an ammeter and a resistance . When the polarity of `E_(2)` is reversed, the current changes from 120 mA to 20 mA . If the combined resistance of ammeter and resistance is `18Omega` then the value of `E_(1)` and `E_(2)` will be

Options

A. 2.5 V, 1.1 V
B. 1.5 V, 2.2 V
C. 1.5 V, 1.1 V
D. 2.5 V, 2.2 V

Answer

C

Explanation

`I=(E_(1)+E_(2))/(R+r_(1)+r_(2)):.E_(1)+E_(2)=2.6` ... (i)
`I=(E_(1)-E_(2))/(R+r_(1)+r_(2)):.E_(1)-E_(2)=0.4` ...(iii)
From equation (i) and (ii),
`E_(1)=1.5V` and `E_(2)=1.1V`