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EMF

In order to d...

Question

In order to determine the emf of a storage battery, it was connected in series with a standard cell in a certain circuit and an current $I_{1}$ was obtained. When the battery is connected to the same circuit opposite to the standard cell a current $I_{2}$ flow in the external circuit from the positive plate of the storage battery. What is the emf $E_{1}$ of the storage battery, if the emf of the standard cell is $E_{2}$ ?

E_{1} = \frac{I_{1} + I_{2}}{I_{1} - I_{2}} E_{2}

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E_{1} = \frac{I_{1} + I_{2}}{I_{2} - I_{1}} E_{2}

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E_{1} = \frac{I_{1} - I_{2}}{I_{1} + I_{2}} E_{2}

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E_{1} = \frac{I_{2} - I_{1}}{I_{1} + I_{2}} E_{2}

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Solution

The correct option is A $E_{1} = \frac{I_{1} + I_{2}}{I_{1} - I_{2}} E_{2}$

Using kirchhoff's loop law, we get:
$- I_{1} r_{2} + E_{2} - I_{1} r_{1} + E_{1} = 0$

$\frac{E_{1} + E_{2}}{r_{1} + r_{2}} = I_{1}$ ....(1)

Again using kirchhoff's loop law:

$E_{1} - I_{2} r_{1} - E_{2} - I_{2} r_{2} = 0$

$\frac{(E_{1} - E_{2})}{r_{1} + r_{2}} = I_{2}$ ....(2)

From equation (1) and (2):

$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{I_{1}}{I_{2}} \Rightarrow E_{1} = \frac{(I_{1} + I_{2})}{I_{1} - I_{2}} E_{2}$

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