In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :

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+ 2V

– 2V

+ 1V

– 1V

answer is C.

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Detailed Solution

Applying KJR at D2−1−i=0 ⇒i=1A Now, applying KVL in branch ACDBVA+1+21−2 =VB⇒VB−VA=1+21−2=+1 volt

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