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

# 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. ${\epsilon }_{1}$ of the storage battery? The e.m.f. of the standard cell is ${\epsilon }_{2}$

1. A

${\mathrm{\epsilon }}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}{{\mathrm{I}}_{1}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{\mathrm{I}}_{2}}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{\epsilon }}_{2}$

2. B

${\mathrm{\epsilon }}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}{{\mathrm{I}}_{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{\mathrm{I}}_{1}}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{\epsilon }}_{2}$

3. C

${\mathrm{\epsilon }}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{\epsilon }}_{2}$

4. D

${\mathrm{\epsilon }}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{I}}_{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{\mathrm{I}}_{1}}{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{\epsilon }}_{2}$

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### Solution:

${\epsilon }_{1}+{\epsilon }_{2}=\text{\hspace{0.17em}}{I}_{1}R,\text{\hspace{0.17em}\hspace{0.17em}}{\epsilon }_{1}-{\epsilon }_{2}={I}_{2}R$

Dividing above two equations, ${\mathrm{\epsilon }}_{1}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\frac{{\mathrm{I}}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{I}}_{2}}{{\mathrm{I}}_{1}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{\mathrm{I}}_{2}}\right){\mathrm{\epsilon }}_{2}$  Register to Get Free Mock Test and Study Material

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