In the figure, ammeters read a current of 1A 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 to be non-ideal and voltmeter to be ideal. The value of resistance R shown in figure in Ω is
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answer is 2.
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Detailed Solution
Let the resistance of ammeter be r Apply KVL for loop HBCIH: - i×1-1×r+3=0⇒i+r=3....... (i) Apply KVL for loop HBCDEFGH: -i×1-1×2r+4=0⇒i+2r=4.......(ii) From (i) & (ii) r=1Ω,i=2A(i−1)R=1×2r⇒R=2Ω