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 I1 was obtained. When polarity of the standard cell is reversed, a current I2. was obtained in the same direction as that of I1. What is thee.m.f. ε1 of the storage battery? The e.m.f. of the standard cell is ε2

ε1+ε2= I1R, ε1−ε2=I2RDividing above two equations, ε1 = I1 + I2I1 − I2ε2

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 I1 was obtained. When polarity of the standard cell is reversed, a current I2. was obtained in the same direction as that of I1. What is thee.m.f. ε1 of the storage battery? The e.m.f. of the standard cell is ε2

ε1+ε2= I1R, ε1−ε2=I2RDividing above two equations, ε1 = I1 + I2I1 − I2ε2

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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 I₁ was obtained. When polarity of the standard cell is reversed, a current I₂. was obtained in the same direction as that of I₁. What is the
e.m.f. $ε_{1}$ of the storage battery? The e.m.f. of the standard cell is $ε_{2}$

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$ε_{1} = \frac{I_{1} + I_{2}}{I_{1} - I_{2}} ε_{2}$

$ε_{1} = \frac{I_{1} + I_{2}}{I_{2} - I_{1}} ε_{2}$

$ε_{1} = \frac{I_{1} + I_{2}}{I_{1} + I_{2}} ε_{2}$

$ε_{1} = \frac{I_{2} - I_{1}}{I_{1} + I_{2}} ε_{2}$

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Detailed Solution

$ε_{1} + ε_{2} = I_{1} R, ε_{1} - ε_{2} = I_{2} R$

Dividing above two equations, $ε_{1} = (\frac{I_{1} + I_{2}}{I_{1} - I_{2}}) ε_{2}$

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