Three batteries having emf E1=12V, E2=14V and E3=2V and internal resistance $R_{1} = 4 Ω$ , $R_{2} = 8 Ω$ and $R_{3} = 12 Ω$ are connected with each other as shown in fig.

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The current through the battery is E1 is 0.5A from left to right.

The potential difference between the points A and B is 10 volt.

The current through the battery is E3 is 1A from left to right.

The internal resistance of the combination across A and B is $\frac{24}{11} Ω$

answer is A, C, D.

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Detailed Solution

$ε_{0} = \frac{(\frac{12}{4} + \frac{14}{8} - \frac{2}{12})}{(\frac{1}{4} + \frac{1}{8} + \frac{1}{12})} = 10 v o l t \frac{1}{4} + \frac{1}{8} + \frac{1}{12} = \frac{1}{R_{0}} R_{0} = \frac{24}{11} Ω$

By the concept of equivalent battery

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