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Question

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


A
+2 V
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B
2 V
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C
1 V
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D
+1 V
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Solution

The correct option is D +1 V

Using Kirchoff's junction at C, we get

i1+i3=i2
1 A+i3=2 Ai3=1 A
Now using Kirchoff's loop law along ACDB
VA+1+i3(2)2=VB
VA++1+i3(2)2=VB
VBVA=32=1 V

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