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. ε1 of the storage battery? The e.m.f. of the standard cell is ε2.

Question

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. ε1 of the storage battery? The e.m.f. of the standard cell is ε2. (A) ε1=(I1+I2/I1-I2) ε2 (B) ε1=(I1+I2/I2-I1) ε2 (C) ε1=(I1-I2/I1+I2) ε2 (D) ε1=(I2-I1/I1+I2) ε2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b981fc861085ab93ab3de2ebcfd3289d-.png" /> ε1+ε2=I1 R, ε1-ε2=I2 R Dividing above two equations, ε1=((I1+I2/I1-I2)) ε2

$ε_{1} = \frac{I _{1} + I _{2}}{I _{1} - I _{2}} ε_{2}$

$ε_{1} = \frac{I _{1} + I _{2}}{I _{2} - I _{1}} ε_{2}$

$ε_{1} = \frac{I _{1} - I _{2}}{I _{1} + I _{2}} ε_{2}$

$ε_{1} = \frac{I _{2} - I _{1}}{I _{1} + I _{2}} ε_{2}$

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

ε1​=I1​−I2​I1​+I2​​ε2​

ε1​=I2​−I1​I1​+I2​​ε2​

ε1​=I1​+I2​I1​−I2​​ε2​

ε1​=I1​+I2​I2​−I1​​ε2​

ε1​+ε2​=I1​R,ε1​−ε2​=I2​R Dividing above two equations, ε1​=(I1​−I2​I1​+I2​​)ε2​

$ε_{1} = \frac{I _{1} + I _{2}}{I _{1} - I _{2}} ε_{2}$

$ε_{1} = \frac{I _{1} + I _{2}}{I _{2} - I _{1}} ε_{2}$

$ε_{1} = \frac{I _{1} - I _{2}}{I _{1} + I _{2}} ε_{2}$

$ε_{1} = \frac{I _{2} - I _{1}}{I _{1} + I _{2}} ε_{2}$

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