The voltage $e_{0}$ in the figure is

48 V

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24 V

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36 V

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28 V

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Solution

The correct option is D 28 V
Applying source conversion

$\frac{e_{0} - 80}{12} + \frac{e_{0}}{12} + \frac{e_{0} - 16}{6} = 0$

$4 e_{0} = 112$

$e_{0} = \frac{112}{4} = 28 V$

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B

E_{0}

The voltage of a cell whose halfcells are given below is:

$M g^{2 +} + 2 e^{-} \to M g (s); E^{0} = - 2.37 V$

$C u^{2 +} + 2 e^{-} \to C u (s); E^{0} = + 0.34 V$ standard EMF of the cell is:

(E_{0})

N i^{+ 2} + 2 e^{-} ⟶ N i; E^{0} = - 0.25 V

Z n^{+ 2} + 2 e^{-} ⟶ Z n; E^{0} = - 0.77 V

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