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In the circuit shown, the reading of the ideal ammeter is 5 A. If the source of emf and the ammeter are interchanged, the reading of the ammeter will be:

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3 A

2 A

5 A

answer is C.

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Detailed Solution

$I = \frac{V (\frac{3 R}{11})}{R + {(\frac{1}{8 R} + \frac{1}{3 R})}^{- 1}}, I^{'} = \frac{V (\frac{3 R}{4})}{8 R + {(\frac{1}{R} + \frac{1}{3 R})}^{- 1}}$

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