question_answer

In order to determine the e.m.f. of a storage battery it was connected in series with a standard cell (both are adding) in a certain circuit and a current \[{{I}_{1}}\] was obtained. When polarity of the standard cell is reversed, a current \[{{I}_{2}}\] was obtained in the same direction as that of \[{{I}_{1}}\] what is the e.m.f. \[{{\varepsilon }_{1}}\] of the storage battery? The e.m.f. of the standard cell is\[{{\varepsilon }_{2}}\].

A) \[{{\varepsilon }_{1}}=\frac{{{I}_{1}}+{{I}_{2}}}{{{I}_{1}}-{{I}_{2}}}{{\varepsilon }_{2}}\]

B) \[{{\varepsilon }_{1}}=\frac{{{I}_{1}}+{{I}_{2}}}{{{I}_{2}}-{{I}_{1}}}{{\varepsilon }_{2}}\]

C) \[{{\varepsilon }_{1}}=\frac{{{I}_{1}}-{{I}_{2}}}{{{I}_{1}}+{{I}_{2}}}{{\varepsilon }_{2}}\]

D) \[{{\varepsilon }_{1}}=\frac{{{I}_{2}}-{{I}_{1}}}{{{I}_{1}}+{{I}_{2}}}{{\varepsilon }_{2}}\]

Correct Answer: A

Solution :

[a] Dividing \[{{\varepsilon }_{1}}=\left( \frac{{{I}_{1}}+{{I}_{2}}}{{{I}_{1}}-{{I}_{2}}} \right){{\varepsilon }_{2}}\]