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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a
+ 2V
b
– 2V
c
+ 1V
d
– 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