In the figure, ammeters read a current of 1 A each as shown, while the voltmeter reads a potential difference of 3 V . The ammeters are identical, the internal resistance of the battery is negligible. Consider all ammeters as non-ideal. The value of resistance R shown in figure in $Ω$ is

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Let the resistance of ammeter be r ​ i × 1 + 1 × r = 3 ⇒ i + r = 3 ​ i × 1 + 1 × 2 r = 4 ⇒ i + 2 r = 4 ( i ) ​ From (i) \& (ii) r = 1Ω , i = 2 A ( i − 1 ) R = 1 × 2 r ⇒ R = 2Ω

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