In the circuit shown, current flowing through 25V cell is

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10A

14.2A

12A

7.2A

answer is C.

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Detailed Solution

The given circuit is
Question Image
As cells in AJ and CH terminal are parallel, their effective emf
$E_{1} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} = \frac{10 \times 5 + 20 \times 5}{10} = \frac{150}{10} = 15 V$
Their effective internal resistance
$r_{1} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{5}{2} = 2 .5 Ω$
Cells in BI and DG are parallel
$\begin{array}{l} ∴ E_{2} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} = \frac{5 \times 11 + 30 \times 10}{10 + 11} \\ = \frac{355}{21} = 16 .9 V \end{array}$
Their effective internal resistance
$r_{2} = \frac{10 \times 11}{10 + 11} = \frac{110}{21} = 5 .28 = 5 .3 Ω$
the circuit now can be drawn as
Question Image
$∴$ The two cells can be replaced by a single cell of
i) $\begin{array}{l} emf E = \frac{E_{1} r_{2} - E_{2} r_{1}}{r_{1} + r_{2}} = \frac{15 \times 5 .3 - 16 .9 \times 2 .5}{7 .8} \\ = \frac{79 .5 - 42 .25}{7 .8} = \frac{37 .25}{7 .8} = 4 .7 V \end{array}$
ii) internal resistance $r = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{2 .5 \times 5 .3}{7 .8}$
= 1.69 = 1.7Ω ∴current through 25V cell is
$i = \frac{E_{eff}}{R_{eff}} = \frac{25 - 4 .7}{1 .7} = 11 .94 ≃ 12 A$